# YouTube Math-Puzzle and Math-Instruction Channels

#### lpetrich

##### Contributor
These YouTube channels have lots of videos showing solutions of mathematics puzzles:
These are more about math exposition and instruction:
Some of the puzzles are problems from various Mathematics Olympiads:
Note: all the problem variables are nonnegative integers.

#### lpetrich

##### Contributor
I'll do all three math puzzles.

2x + 3y = n2

Do modular arithmetic, modulo 4:

x: 2x -> 0: 1, 1: 2, (2: 0)
y: 3y -> (0: 1, 1: 3)
n: n2 -> (0: 0, 1: 1, 2: 0, 3: 1)
The () means repeats

So x = 0, y even, n even, x = 1, y odd, n odd, or x = 2, y even, n even

For x = 0, take y = 2y' and rearrange:
1 = n2 - 32y' = (n - 3y')(n + 3y')
No solution.

For x = 1, try mod 8:
y: 3y -> (0: 1, 1: 3)
n: n2 -> (0: 0, 1: 1, 2: 4, 3: 1, 4: 0, 5: 1, 6: 4, 7: 1)
The lhs is 3 or 5, while the rhs is 0, 1, or 4.
No solution.

For x = 2, try mod 8 also.
The lhs is 5 or 7.
No solution.

So we have x > 2.
For a solution, y must be even. Take y = 2y' and rearrange:
2x = n2 - 32y' = (n - 3y')(n + 3y')

3x + 4y = 5z

2x - 3y = 7

Solution:

#### lpetrich

##### Contributor
2x + 3y = n2

Do modular arithmetic, modulo 4:

x: 2x -> 0: 1, 1: 2, (2: 0)
y: 3y -> (0: 1, 1: 3)
n: n2 -> (0: 0, 1: 1, 2: 0, 3: 1)
The () means repeats

So x = 0, y even, n even, x = 1, y odd, n odd, or x = 2, y even, n even

For x = 0, take y = 2y' and rearrange:
1 = n2 - 32y' = (n - 3y')(n + 3y')
No solution.

For x = 1, try mod 8:
y: 3y -> (0: 1, 1: 3)
n: n2 -> (0: 0, 1: 1, 2: 4, 3: 1, 4: 0, 5: 1, 6: 4, 7: 1)
The lhs is 3 or 5, while the rhs is 0, 1, or 4.
No solution.

For x = 2, try mod 8 also.
The lhs is 5 or 7.
No solution.

So we have x > 2.
For a solution, y must be even. Take y = 2y' and rearrange:
2x = n2 - 32y' = (n - 3y')(n + 3y')

Let x = x1 + x2
n = (2x1 + 2x2)/2
3y' = (-2x1 + 2x2)/2

At least one of x1 and x2 must be > 0, and for divisibility by 2, both of them must be: x1 = x1' + 1, x2 = x2' + 1
x = x1' + x2' + 2
n = 2x1' + 2x2'
3y' = -2x1' + 2x2'

For the last one, x1' must be zero and x2' nonzero to make an odd number:
x = x2' + 2
n = 1+ 2x2'
3y' = -1+ 2x2'

Take mod 8 on the last one.
lhs: 1, 3
x2': rhs: 0: 7, 1: 1, 2: 3, (3: 7)

That gives two solutions: x2' = 1 y' = 0, and x2' = 2 y' = 1.

23 + 30 = 32 -- 8 + 1 = 9
24 + 32 = 52 -- 16 + 9 = 25

#### lpetrich

##### Contributor
3x + 4y = 5z

First, try mod 4:
x: 3x -> (0: 1, 1: 3)
y: 4y -> 0: 1, (1: 0)
5z = 1

So x must be even and y > 0. Try mod 8:
x: 3x -> (0: 1, 1: 3)
y: 4y -> 0: 1, 1: 4, (2: 0)
z: 5z -> (0: 1, 1: 5)

So y = 1 and z is odd, or y > 1 and z is even. Try mod 3:

x: 3x -> 0: 1, (1: 0)
y: 4y -> (0: 1)
z: 5z -> (0: 1, 1: 2)

So x = 0 and z is odd, or x > 0 and z is even. From the previous result, if z is odd, then y = 1, and we have a solution:

30 + 41 = 51 -- 1 + 4 = 5

For the rest of the solutions, x and z must be even and > 0. Rearranging,

22y = 52z' - 32x' = (5z' - 3x') (5z' + 3x')

Setting 2y = (1 + y1) + (1 + y2),
we find
3x' = - 2y1 + 2y2
5z' = 2y1 + 2y2

Since 3 and 5 are odd, y1 must be 0, and y2 must be even. Setting it to 2y':
2y = 2 + y2
3x' = - 1 + 22y'
5z' = 1 + 22y'

Taking mod 8, 3x' is 1 or 3, and its rhs is y' = 1: 3, y' >= 2: 7. So y' = 1, x' = 1, z' = 1, giving the only other solution:

32 + 42 = 52 -- 16 + 9 = 25

#### lpetrich

##### Contributor
2x - 3y = 7

First, since 2x > 7, x >= 3.

Using modular arithmetic, with mod 4, we find that y is even. Likewise, with mod 3, either x is odd and y = 0, or else x is even and y > 0.

For y = 0, we find 2x = 7 + 1 = 8 = 23, and we have a solution:

23 - 30 = 8 - 1 = 7

For y = 0, both x and y are even (x = 2x', y = 2y'), and we have

(2x' - 3y') (2x' + 3y') = 7

We are helped by 7 being a prime number: 2x' - 3y' = 1 and 2x' + 3y' = 7 giving us 2x' = 4 and 3y' = 3, or x' = 2 and y' = 1. This gives us the remaining solution:

24 - 32 = 16 - 9 = 7

#### lpetrich

##### Contributor
More:

Here is one of SyberMath's puzzles. Solve for x:

$$\displaystyle{ \frac{(x+1)^5}{x^5+1} = \frac{81}{11} }$$

Banned
42

#### lpetrich

##### Contributor
I don't recall where I got this one, but I must state its solution:

2a + 3b + 5c = n!

where a, b, c, and n are nonnegative integers.

This means that n! >= 3, meaning that n >= 2.

By trial and error, one can show that (a,b,c,n) has values (1,1,0,3), (2,0,0,3), and (4,1,1,4). Are there any other solutions?

We can find constraints by doing modular arithmetic. Let us take n >= 5 for simplicity, since we have found all the solutions for n <= 4.

First, modulo 2:
a: 1, (0)
b: (1)
c: (1)
The () means repeated. Solutions: a >= 1

Modulo 4:
a: 1, 2, (0)
b: (1, 3)
c: (1)
Solutions: a = 1, b even / a > 1, b odd

Modulo 8:
a: 1, 2, 4, (0)
b: (1, 3)
c: (1, 5)
Solutions: a = 1, b even, c odd / a = 2, b odd, c even / a >= 3, b odd, c odd

Modulo 3:
a: (1, 2)
b: 1, (0)
c: (1, 2)
Sollutions: b = 0, a even, c even / b >= 1, a even, c odd / b >= 1, a odd, c even

Modulo 5:
a: (1, 2, 4, 3)
b: (1, 3, 4, 2)
c: 1, (0)
Solutions:
c = 0, a = 0m4, b = 1m4 / c = 0, a = 1m4, b = 3m4 / c = 0, a = 3m4, b=0m4
c >= 1, a = 0m4, b = 2m4 / c >= 1, a = 1m4, b = 1m4 / c >= 1, a = 2m4, b = 0m4 / c >= 1, a = 3m4, b = 3m4

Let us consider c = 0 first. By mod 8, b is odd and a = 2. By mod 3, b > 1 and a is odd. Contradiction in the value of a.

Now c > 0. It is a special case of c > 0, a even, b even / c > 0, a odd, b odd.

Combining with mod 3, we have a even, b = 0, c even >= 1 / a even, b even >=1, c odd >= 1 / a odd, b odd >= 1, c even >= 1

Of the mod 8 solutions, #1 a = 1, b even, c odd agrees with the second one except for a. #2 a = 2, b odd, c even agrees with the third one except for a, #3 a >= 3, b odd, c odd agrees with none of them.

Thus, by contradiction, there are no solutions for n >= 5, and the three previously-found solutions are all of them.

#### Swammerdami

Staff member
I like these videos, especially Mathologer. One very good puzzle is to view the Numberphile video proving that All Triangles are Equilateral and find the fallacy yourself. (SSS, SAS, ASA are the three usual ways to prove triangle congruence. The proof uses an SSA theorem: Is that the fallacy?)

Here are two historical results from the 14th century related to the types of equation Ipetrich is writing about:

(a) xa - yb = 1 has only one solution in integers greater than 1. Catalan's conjecture
The only solution is 32 - 23 = 1. The Conjecture wasn't proven until the 21st century, but a special case (restricting (x,y) to (2,3) or (3,2)) was proven in the 14th century by Levi ben Gerson. Can you prove that special case?

(b) x4 + y4 = z4 has no solutions in positive integers
This special case (n=4) of FLT is much easier to prove than the n=3 case (though still rather difficult), and — little-acknowledged fact — was effectively proven in the 14th century by the great Leonardo of Pisa.

#### lpetrich

##### Contributor
I will do so for the 2-3 case. I'll take on the two together as 2a - 3b = +-1.

Four solutions fpr a <= 3:
(1,0,+1), (1,1,-1), (2,1,+1), (3,2,-1)

Mod 8:
2: 1, 2, 4, (0)
3: (1, 3)

For a >= 3, only -1 is possible, and b must be even.

This means that b = 2b' and 2a = 32b' - 1 = (3b' - 1) (3 b' + 1)

Let a = a1 + a2. We can then decompose the equation as a product of the equations 2a1 = 3b' - 1 and 2a2 = 3b' + 1. Subtracting the first one from the second one gives 2a2 - 2a1 = 2.

That gives us a1 >= 1 and 2a2 = 2 * (2a1-1 + 1) and then 2a2-1 = 2a1-1 + 1

There is only one possible solution: a1 = 1 and a2 = 2, giving a = 3 and then b = 2 and the difference equaling -1.

So the above four solutions are the only ones, and for a, b >= 2, there is only one: (3,2,-1).

#### none

##### Banned
Banned
lpetrich
I made a gallery in javascript in 14 tries.... then it was clear of syntax errors.
8 images, well it is indefinite... the limit is as much memory my computer could handle without crashing... at the time, that is.
even or odd.
it's a script... no conditional statements
just wait for the "click" redirect from a trig function. no "if... then" just a counter derived from the number of images in the .html
the backend was well you know a primitive berkley script convert.... you know magick...
any uploaded via ftp called by cron to intercept and apply the water marks...
algebra is great... y=mx+b
fluid dynamics is next, well that was a decade ago..
deriving the ability to y = mx + b from an indeterminate is next on my plate.
but yeah. ic
stiil i find a=1, b=1, c=1 , versus
a=1
b=1
c=1
is less intelligible.
then there is football and basketball, you know put the sphere in the square hole... because it fits.

#### lpetrich

##### Contributor
Michael Penn has this puzzle:

He starts with 297 = 25 + 27 + ... 39 + 41 and he asks which numbers can be expressed as the sum of consecutive odd numbers. Can 2022 be expressed in that way?

Michael Penn's own proof for the problem 2a + 3b + 5c = n! uses mod 120, thus incorporating mod 8, mod 3, and mod 5.

a: 1, 2, 4, (8, 16, 32, 64)
b: 1, (3, 9, 27, 81)
c: 1, (5, 25)

One then finds which selections of them add up to 0 mod 120, and one finds none. Thus n <= 4, and the three solutions found earlier are the only ones.

Now this SyberMath puzzle:
$$\displaystyle{ \frac{(x+1)^5}{x^5+1} = \frac{81}{11} }$$

Take LHS - RHS and multiply by the LCM of the denominators:
$$11 (x+1)^5 - 81 (x^5+1) = - 5 (14 x^5 - 11 x^4 - 22 x^3 - 22 x^2 - 11 x+ 14)$$

This polynomial has the property that if x is a solution, then 1/x is also a solution. If x and 1/x are not distinct, then x = 1 or -1. We find that x = -1 is a root, and we thus divide out (x + 1). That gives us
$$(x + 1) (14 x^4 - 25 x^3 + 3 x^2 - 25 x + 14)$$

x = -1 gives 0/0 = 81/11 in the original expression, so we leave that one aside.

The rational roots of the fourth-order polynomial have form
$$\displaystyle{ x = \pm \frac{ \text{some factor of 14} }{ \text{some factor of 14} } }$$

14 has factors 1, 2, 7, and 14, including itself, and one can then try out these possibilities. The only solutions are x = 2 and x = 1/2, and we find
$$(x - 2)(2x - 1)(7 x^2 + 5 x + 7)$$

$$\displaystyle{ x = \frac{ - 5 \pm 3 i \sqrt{19} } {14} }$$

So the only real solutions of the original problem are the rational ones, x = 2 and x = 1/2.

#### lpetrich

##### Contributor
More:

What primes p satisfy $$p^3 - 4p + 9 = n^2$$ for nonnegative integer n?

What positive integers m, n satisfy $$n^2 + 2021 n = m^2$$ ?

What is $$\sqrt{ 3 \sqrt{ 5 \sqrt{ 3 \sqrt{ 5 \dots } } } }$$ ?

What real x satisfies $$\sqrt{x} - \sqrt{x} = \sqrt{x} - \sqrt{x}$$ ?

For every prime p, is there an n such that $$2^n + 3^n + 6^n - 1$$ is a multiple of p?

What real roots (x,y,z) are there of
$$(x - 1) (y - 1) (z - 1) = x y z - 1$$
$$(x - 2) (y - 2) (z - 2) = x y z - 2$$
?

What primes p, q satisfy $$p q | (5^p - 2^p) (5^q - 2^q)$$ ?

Is this an integer? $$4\sqrt{4 - 2\sqrt{3}} + \sqrt{97 - 56\sqrt{3}}$$

Find this sum:
$$\displaystyle{ \sum_{n=2}^\infty \frac{n^4 + n^3 + n^2 - n + 1}{n^6 - 1} }$$

#### lpetrich

##### Contributor
Another Michael Penn puzzle: find x, y in N0 (nonnegative integers) such that 7x + 2 = 3y

Solutions to some previous puzzles.

n2 + 2021*n = m2

Do it generally: n2 + a*n = m2

For a even, we can take a -> 2a, and n2 + 2a*n = m2

Completing the square and rearranging, one gets (n + a)2 - m2 = a2

Now take a2 = a1*a2, factor the left-hand side, and set the two equal:
n + a - m = a1
n + a + m = a2
One gets n = (a1+a2)/2 - a and m = (-a1+a2)/2

For general a in the first one, multiply by 4, complete the square, rearrange, and factor:
(2n + a + 2m) * (2n + a - 2m) = a2

This gives us n = (a1+a2)/4 - a/2 and m = (-a1+a2)/4

Considering a1 = 1 and a2 = a2 and related factorizations,

n = ((a-1)/2)2 or n = - ((a+1)/2)2 and m = +- (a2-1)/4

Also, a1 = a2 = a gives n = 0 or -a, m = 0

Since a = 2021 = 43*47 here, we get these solutions:
(1,20212)
(432,472)
(43*47,43*47)
n = 1020100 or -1022121 and m = +- 1021110
n = 4 or -2025 and m = +- 90
n = 0 or -2021, m = 0

Last edited:

#### lpetrich

##### Contributor
Yet another problem. A problem with "hidden twin primes".
$$\displaystyle{ \frac{p}{p+1} + \frac{q+1}{q} = \frac{2n}{n+2} }$$

Back to solutions.

Setting x to that repeated square root, one finds that it satisfies $$x = \sqrt{ 3 \sqrt{ 5 x } }$$

Expanding out in powers gives $$x = 3^{1/2} 5^{1/4} x^{1/4}$$

It is easy to solve for x, and one gets $$x = 3^{2/3} 5^{1/3}$$

$$\sqrt{x} - \sqrt{x} = \sqrt{x} - \sqrt{x}$$
is equivalent to
$$x^{1/7} - x^{1/5} = x^{1/3} - x^{1/2}$$

The denominators have least common multiple 210, and one can set x = y210. That gives us
y30 - y42 - y70 + y105

The rational roots are 0 (multiplicity 30) and 1. The remaining polynomial, with degree 74, could not be factored by Mathematica, though it has two real roots, one positive and one negative.

I may have misremembered it. Here is a simpler problem:
$$\sqrt{x} - \sqrt{x} = \sqrt{x} - x$$
That is equivalent to
$$x^{1/7} - x^{1/5} = x^{1/3} - x$$

The denominators' least common multiple is 105, and one can set x = y105. That gives us
y15 - y21 - y35 + y105

The rational roots are 0 (multiplicity 15) and +-1. There are complex rational roots, +-i. Likewise, the remaining polynomial, with degree 74, could not be factored by Mathematica, though it also has two real roots, one positive and one negative.

Last edited by a moderator:

#### lpetrich

##### Contributor
Let us solve
$$4\sqrt{4 - 2\sqrt{3}} + \sqrt{97 - 56\sqrt{3}}$$
To do that, let us note that $$(a + b \sqrt{c})^2 = (a^2 + b^2 c) + 2ab \sqrt{c}$$

For the first one, a*b = - 1, meaning a = 1 and b = 1, or a = -1 or b = 1. Selecting the signs that give a positive value, $$(\sqrt{3} - 1)^2 = 4 - 2\sqrt{3}$$

For the second one, a*b = - 28, and since (a2 + b22 * c) must be odd, one or the other of a must be odd. That means a = 4, b = -7 (or reversed signs) or a = 7, b = -4 (or reversed signs). The first one produced the right value, and selecting the signs that give a positive value, $$(7 - 4 \sqrt{3})^2 = 97 - 56 \sqrt{3}$$

Thus we get overall $$4 (\sqrt{3} - 1) + (7 - 4 \sqrt{3}) = 3$$

Turning to
(x-1)*(y-1)*(z-1) = x*y*z-1
(x-2)*(y-2)*(z-2) = x*y*z-2

Expanding the two, we get
- (x*y + x*z + y*z) + (x + y + z) = 0
- (x*y + x*z + y*z) + 2*(x + y + z) = 3

That gives us
(x + y + z) = 3
(x*y + x*z + y*z) = 3
Squaring the first one and subtracting 2 * the second one, we get
(x2 + y2 + z2) = 3

Subtracting 2 * the first one and adding 3, we get
(x - 1)3 + (y - 1)3 + (z - 1)3 = 0

The only real solution is thus x = y = z = 1.

Find this sum:
$$\displaystyle{ \sum_{n=2}^\infty \frac{n^4 + n^3 + n^2 - n + 1}{n^6 - 1} }$$

The denominator can be factored: $$n^6 - 1 = (n - 1) (n + 1) (n^2 - n + 1) (n^2 + n + 1)$$

So we now break apart this expression into expressions with each of the denominator's factors:
$$\displaystyle{ \frac12 \sum_{n=2}^\infty \left( \frac{1}{n - 1} - \frac{1}{n + 1} + \frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1} \right) }$$

The second term is the first term shifted by n -> n+2, and the fourth term the third term shifted by n -> n+1. That gives us
$$\displaystyle{ \frac12 \left( \sum_{n=2}^3 \frac{1}{n - 1} + \sum_{n=2}^2 \frac{1}{n^2 - n + 1} \right) = \frac12 \left( 1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{11}{12} }$$

#### lpetrich

##### Contributor
Oops about my solution of a previous problem. Its exponents are 2 not 3:
(x - 1)2 + (y - 1)2 + (z - 1)2 = 0

Another Michael Penn problem. Solve this ordinary differential equation:
$$\displaystyle{ \frac{dy}{dx} + \frac{y}{x} = x \sqrt{y} }$$

Now some problems from Let's Think Critically:

Find
$$\displaystyle{ \sqrt{7 - \sqrt{7 + \sqrt{7 - \sqrt{7 + \dots} } } } }$$

For variables x, y, z:
$$x^2 y + y^2 z + z^2 x = 2186 \\ x^2 z + y^2 x + z^2 y = 2188 \\ x^2 + y^2 + z^2 = \text{ ?}$$

Find
$$\displaystyle{ 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \dots }$$

$$\displaystyle{ \prod_{k=2}^\infty \frac{k^3 - 1}{k^3 + 1} }$$

What is the second expression as a function of a?
$$\displaystyle{ \frac{x}{x^2 + x + 1} = a ,\ \frac{x^2}{x^4 + x^2 + 1} = \text{ ?} }$$

What are x, y, z?
$$x + y + z = 3 \\ x^3 + y^3 + z^3 = 15 \\ x^5 + y^5 + z^5 = 83$$

#### lpetrich

##### Contributor
I will now try to solve
$$\displaystyle{ \frac{p}{p+1} + \frac{q+1}{q} = \frac{2n}{n+2} }$$

for primes p, q and integer n. Rearranging and merging the fractions, I find
$$(1 + p + q + 2 p q) + n (1 + p - q)$$

For q = 2,
$$\displaystyle{ n = \frac{3 + 5p}{1 - p} = - 5 + \frac{8}{1 - p} }$$

The solutions are p = 2, 3, 5

For p = 2,
$$\displaystyle{ n = \frac{3 + 5q}{-3 + q} = 5 + \frac{18}{-3 + q} }$$

The solutions are q = 2, 5

The combined solution for one prime equal to 2 are (2,2,-13), (2,5,14), (3,2,-9), (5,2,-7)

We now turn to the case of both primes being odd. In general,
$$\displaystyle{ n = - \frac{1 + p + q + 2 p q}{1 + p - q} = - 1 - 2q - \frac{2 q^2}{1 + p - q} }$$

The denominator is always odd, so it cannot divide 2, and must therefore divide q. Solve for p in 1+p-q = ...
• +1: p = q
• -1: p = q - 2
• +q: p = 2*q - 1
• -q: p = -1 -- not possible
with solutions
(q, q, -(2q2+2q+1)), (q-2, q, (2q2-2q-1)), (2q-1, q, -(4q+1))

where q-2 must be a prime in the second one and 2q-1 a prime in the third one.

The second one thus needs twin primes, and the third one needs the primes in A005382 - OEIS - "Primes p such that 2p-1 is also prime."

#### lpetrich

##### Contributor
Now, for prime p and nonnegative integer n,
p3 - 4*p + 9 = n2

There is one solution: p = 2, n = 3.

All the other primes are odd, and of these, p = 3 is not a solution. So we consider p > 3.

Take modulo p. That gives n2 = 9 mod p or n = 3 + k*p

Inserting, subtracting 9, and dividing by p,
(p - 2) * (p + 2) = k*(k*p + 6)

p has forms 3*m+1 or 3*m+2. Both of them will make (p2 - 4) divisible by 3, and that means that k is also divisible by 3. Thus, we can take n = 3*(k*p + 1) giving us
(p - 2) * (p + 2) = 9*k*(k*p + 2)

Let us solve for p:
$$\displaystyle{ p = \frac{1}{2} \left( 9 k^2 \pm \sqrt{81 k^4 + 72 k + 16} \right) }$$

Using the first term in the square root gives 9*k2, and since the actual expression in the square root is different, we get these constraints for it being an integer:
• 81*k4 + 72*k + 16 >= (9*k2 + 1)2
• 81*k4 + 72*k + 16 <= (9*k2 - 1)2
They simplify to
• 72*k + 15 >= 18*k2
• 72*k + 15 <= - 18*k2
and we get -3 <= k <= 4. Trying that range of k values into the expression for p gives positive-integer values 2, 7, 11. So we have solutions
(2,3), (7,18), (11,36)

#### lpetrich

##### Contributor
From Let's Think Critically: nonnegative integers a, b, n with 2a + 3b = n2

Solve for y as a function of x:
$$\displaystyle{ \frac{dy}{dx} + \frac{y}{x} = x \sqrt{y} }$$

Set y = u2 and divide the expression by 2*u:
$$\displaystyle{ \frac{du}{dx} + \frac{u}{2x} = \frac{x}{2} }$$

This is an easy kind of differential equation to solve. Let us consider a general version:
$$\displaystyle{ \frac{dy}{dx} + f(x) y = g(x) }$$

Solve the homogeneous version first, the version with g(x) = 0. That gives us
$$\displaystyle{ y = \exp (-F(x) ) ,\ F(x) = \int f(x) \, dx }$$

To find the general solution, take
$$y = z(x) \exp (-F(x) )$$

and one finds
$$\displaystyle{ \frac{dz}{dx} = g(x) \exp (F(x) ) }$$

which is easy to solve. Putting the pieces together,
$$\displaystyle{ y = \exp (-F(x)) \left( C + \int \exp (F(x)) g(x) \, dx \right) }$$

where C is a constant of integration. In our current problem, f(x) = 1/(2*x) and g(x) = x/2, and we find F(x) = (1/2)*log(x) and
$$\displaystyle{ u = C x^{-1/2} + \frac15 x^2 }$$

#### lpetrich

##### Contributor
Find
$$\displaystyle{ \sqrt{7 - \sqrt{7 + \sqrt{7 - \sqrt{7 + \dots} } } } }$$

This is x in
$$\displaystyle{ x = \sqrt{7 - \sqrt{7 + x} } }$$

Square it:
$$\displaystyle{ x^2 = 7 - \sqrt{7 + x} }$$

Subtract 7 and square again:
$$\displaystyle{ (x^2-7)^2 = 7 + x }$$

This expands into
$$x^4 - 14 x^2 - x + 42$$

One can factor this polynomial by finding which gives zero out of everything that evenly divides 42, using both positive and negative signs. This gives x = 2 and x = -3. Thus
$$x^4 - 14 x^2 - x + 42 = (x - 2) (x + 3) (x^2 - x - 7)$$

$$\displaystyle{ x = \frac12 \left( 1 \pm \sqrt{29} \right) }$$

The possible solutions to the original equation must have all square roots positive. This rules out the two negative solutions, so let us consider the remaining two. The one with sqrt(29) gives the wrong value, which means that that solution requires a negative sign for the inner square root. The remaining solution is x = 2:

$$\displaystyle{ \sqrt{7 - \sqrt{7 + 2} } = \sqrt{7 - \sqrt{9} } = \sqrt{7 - 3} = \sqrt{4} = 2 }$$

#### lpetrich

##### Contributor
What is the second expression as a function of a?
$$\displaystyle{ \frac{x}{x^2 + x + 1} = a ,\ \frac{x^2}{x^4 + x^2 + 1} = \text{ ?} }$$

Let the second one equal b. Multiply by the denominators and find the polynomial equations:
$$a x^2 + (a - 1) x + a = 0 \\ b x^4 + (b - 1) x^2 + b = 0$$

One can avoid directly solving for x by finding the remainder after dividing the second polynomial by the first one. That gives us
$$\displaystyle{ - \frac{1}{a^3} (2 a b - b + a^2) (a x + a - x) }$$

One can solve for x, and one finds that only a = 0 and x = 0 make that solution possible. For nonzero a, we must solve for b:
$$\displaystyle{ b = \frac{a^2}{1 - a} }$$

What are x, y, z?
$$x + y + z = 3 \\ x^3 + y^3 + z^3 = 15 \\ x^5 + y^5 + z^5 = 83$$

Solve the first equation for z and plug the solution into the other two equations. The second one becomes a quadratic equation in x and y and the second one a quartic one (degree 4 in each variable). Take the polynomial remainder of the third one relative to the second one, treating them as polynomials in y, and one gets
$$- 20 (x - 1) (x^2 -2x - 1)$$
with solutions
$$x = 1 ,\ x = 1 + \sqrt{2} ,\ x = 1 - \sqrt{2}$$

So the solutions of the original equations are all six permutations of (variable) = (value), like
$$x = 1 ,\ y = 1 + \sqrt{2} ,\ z = 1 - \sqrt{2}$$

#### lpetrich

##### Contributor
Find
$$\displaystyle{ 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \dots }$$

This has a close resemblance to
$$\displaystyle{ \log(1 + x) = \sum_{k=0}^\infty \frac{(-1)^k x^{k+1}}{k+1} = x - \frac12 x^2 + \frac13 x^3 - \frac14 x^4 + \dots }$$

The series 1, 4, 7, 11, ..., is every third term in the above expression after the first one, for x = 1. This suggests that we add over the cube roots of unity. For powers 1 and 2 modulo 3, that sum is zero, while for a multiple-of-3 power, that sum is 3. Thus,
$$\displaystyle{ \frac13 \sum_{x | x^3 = 1} \frac{1}{x} \log(1 + x) = \sum_{k=0}^\infty \frac{(-1)^k }{3k+1} = 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \dots }$$

The cube roots of 1 are
$$1 ,\ \frac12 ( - 1 + i \sqrt{3} ) ,\ \frac12 ( - 1 + i \sqrt{3} )$$

and the above expression becomes
$$\frac13 \left( \log(2) + \frac12 ( - 1 - i \sqrt{3} ) \log( \frac12 (1 + i \sqrt{3} ) ) + \frac12 ( - 1 + i \sqrt{3} ) \log( \frac12 (1 - i \sqrt{3} ) ) \right) \\ = \frac13 \left( \log(2) + \sqrt{3} \arctan(\sqrt{3}) \right) \\ = \frac13 \left( \log(2) + \pi / \sqrt{3} \right)$$

For positive integers x, y, z:
$$x^2 y + y^2 z + z^2 x = 2186 \\ x^2 z + y^2 x + z^2 y = 2188 \\ x^2 + y^2 + z^2 = \text{ ?}$$

If x, y, z are all even, then the lhs's must be divisible by 8, something true of neither 2186 nor 2188. If x, y, z are all odd, then they add up to an odd number, and likewise if one of them is even. Thus, two must be even and one odd. Let us take the even ones to be y and z: y -> 2y, z -> 2z
$$2 x^2 y + 8 y^2 z + 4 z^2 x = 2186 \\ 2 x^2 z + 4 y^2 x + 8 z^2 y = 2188$$

Divide both equations by 2:
$$x^2 y + 4 y^2 z + 2 z^2 x = 1093 \\ x^2 z + 2 y^2 x + 4 z^2 y = 1094$$

Repeat for z: z -> 2z and dividing the 2nd eqn by 2
$$x^2 y + 8 y^2 z + 8 z^2 x = 1093 \\ x^2 z + y^2 x + 8 z^2 y = 547$$

We find that x and y must be odd. Repeat for z: z -> 2z
$$x^2 y + 16 y^2 z + 32 z^2 x = 1093 \\ 2 x^2 z + y^2 x + 32 z^2 y = 547$$

Taking x -> 2x+1 and y -> 2y+1, subtracting 1 and dividing by 2:
$$4 x^2 y + 32 y^2 z + 32 z^2 x + 2 x^2 + 16 z^2 + 4 x y + 32 y z + 2 x + y + 8 z = 546 \\ 4 x^2 z + 4 y^2 x + 32 z^2 y + 2 y^2 + 16 z^2 + 4 x y + 4 x z + x + 2 y + z = 273$$

Taking y -> 2y and dividing the 1st eqn by 2
$$4 x^2 y + 64 y^2 z + 16 z^2 x + x^2 + 8 z^2 + 4 x y + 32 y z + x + y + 4 z = 273 \\ 4 x^2 z + 16 y^2 x + 64 z^2 y + 8 y^2 + 16 z^2 + 8 x y + 4 x z + x + 4 y + z = 273$$

This means that y must be odd. Considering the term 64*y^2*z, the only value of y that permits it to be less than 273 is y = 1. Undoing the substitutions gives 1, 2, 5, 10. So y = 10 and the equations are now
$$z^2 x + 10 x^2 + 100 z = 2186 \\ x^2 z + 10 z^2 + 100 x = 2188$$

Repeat z -> 8z and divide by 2 and 4:
$$32 z^2 x + 5 x^2 + 400 z = 1093 \\ 2 x^2 z + 160 z^2 + 25 x = 547$$

The term 160*z^2 must be less than 547, and the only possible value of z is 1. Undoing the substitutions gives z = 8.

So we find
$$10 x^2 + 64 x = 1386 \\ 8 x^2 + 100 x = 1548$$

One can eliminate the quadratic term by taking 4 * (first equation) and subtracting 5 * (second equation). That gives us 244*x = 2196, with solution x = 9. One can easily show that this satisfies both of the equations.

So we have x = 9, y = 10, z = 8, or some rotation of the three values, with x^2 + y^2 + z^2 = 245.

#### lpetrich

##### Contributor
Now we solve
$$\displaystyle{ \prod_{k=2}^\infty \frac{k^3 - 1}{k^3 + 1} }$$

Factor the numerator and the denominator and make the product finite:
$$\displaystyle{ \prod_{k=2}^N \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} }$$

Break it up:
$$\displaystyle{ \prod_{k=2}^N (k-1) \prod_{k=2}^N \frac{1}{k+1} \prod_{k=2}^N (k^2+k+1) \prod_{k=2}^N \frac{1}{k^2-k+1} }$$

$$\displaystyle{ \prod_{k=2}^N (k-1) \prod_{k=4}^{N+2} \frac{1}{k-1} \prod_{k=3}^{N+1} (k^2-k+1) \prod_{k=2}^N \frac{1}{k^2-k+1} }$$

Most of these summations now cancel, and for large N, one finds

$$\displaystyle{ \frac{2}{N^2} \frac{N^2}{3} = \frac{2}{3} }$$

Thus,
$$\displaystyle{ \prod_{k=2}^\infty \frac{k^3 - 1}{k^3 + 1} = \frac23 }$$

#### lpetrich

##### Contributor
I'll try 2a + 3b = n2 for nonnegative integers 2, 3, n.

Let's first try modulo 3:
• 2: (1, 2)
• 3: 1, (0)
• square: 0: 0, 1: 1, 2: 1 (only values: 0, 1)
For b = 0, a must be odd, and for b > 0, a must be even.

Let us consider b = 0. Then 2a + 1 = n2 Subtracting 1 gives
2a = n2 - 1 = (n - 1) * (n + 1)

Splitting into factors,
2a2 = n + 1 and 2a1 = n - 1
where a1 and a2 are nonnegative integers that add to make a. Some rearrangement gives us
2a2 - 2a1 = 2
This means that a2 > a1 and a1 > 0, and
(2a2-a1 - 1) 2a1-1 = 1
giving a1 = 1 and
2a2-a1 -1 = 1 then 2a2-a1 = 2
giving a2 = 2.Thus, a = 3:
23 + 30 = 8 + 1 = 9 = 32

Turning to b > 0, a is even, and I will set it to 2a'. Then, 22a' + 3b = n2 giving
3b = n2 - 22a' = (n + 2a') * (n - 2a')

Splitting 3b like 2a earlier,
3b2 = n + 2a' and 3b1 = n - 2a'
giving
2a'+1 = 3b1 (3b2-b1 - 1)

This means that b1 = 0 and b2 = b: 2a'+1 = 3b - 1

Taking mod 3 gives us even a', a' = 2a''. Taking mod 4 gives us a' > 0 -> b even = 2b'. For a' = 0, we find b = 1, and
20 + 31 = 1 + 3 = 4 = 22

22a''+1 = 32b' - 1 = (3b' - 1) * (3b' + 1)
Splitting the power of 2 as 2a''+1 = a1 + a2,
2a2 = 3b' + 1 and 2a1 = 3b' - 1
Doing the calculation done earlier, a2 = 2, a1 = 1, a'' = 1, a' = 2, a = 4, and b = 3:
24 + 32 = 16 + 9 = 25 = 52

The three solutions (a,b,n): (0,1,2), (3,0,3), (4,2,5)

#### lpetrich

##### Contributor
Another Michael Penn problem:
$$\displaystyle{ \frac{n^3+3}{n^2+7} = m }$$

where both n and m are positive integers. I'll do it more generally for n and m integers that may be zero or negative.

I earlier misuderstood it as
$$\displaystyle{ \frac{n^3+7}{n^2+3} = m }$$

Subtracting n from both sides gives
$$\displaystyle{ \frac{-7n+3}{n^2+7} = m - n = m' }$$

$$\displaystyle{ \frac{-3n+7}{n^2+3} = m - n = m' }$$

In both cases, since the LHS tends to zero as n tends to infinity in both directions, that means that one only needs to test a finite number of n's to find all solutions. When one does that, one gets (n,m) = (2,1), (5,4) for the first one and n = (-4,-3), (1,2) for the second one. The first one's solutions are all positive, while only the second one's second solution is all positive.

#### Swammerdami

Staff member
Here's one of my favorite geometric proofs found at YouTube. The theorem is that all triangles are isoceles. This marvelous theorem has all sorts of surprising corollaries. Euclid's Books have a large number of theorems, some with very clever proofs. But Euclid missed this one!

I've selected points A, B and C in general position in the plane, I will prove that AB = AC, i.e. that the triangle ABC (shown in yellow) is isoceles.

I drew this diagram quickly and free-hand, but geometric proofs don't depend on the accuracy of a drawing. Geometric proofs just rely on simpler theorems. The only theorems we will need are the SAS, ASA and SSA theorems of triangle congruence. (SSA congruence does not hold in general, but it does hold for many special cases, such as whenever the angle is a right angle as it is each time we apply SSA.)

We have constructed a bisector of the angle BAC. We have also bisected the segment BC and drawn a perpendicular from its midpoint. That perpendicular bisector and the angle bisector meet at a point, which we've labeled E. (If DE and AE are the same line choose any point along that line for E.)

Now, from E drop a perpendicular to AB; call the intersection F. Similarly, G is the point on AC which forms a perpendicular with E. We've shown BD=CD (constructed), BE=CE (SAS), and EF=EG (SSA).

Now CGE and BFE are congruent triangles (SSA) so CG=BF. And AGE and AFE are congruent (AAS) so AG=AF. Subtracting equals from equals (AG - CG) = (AF - BF). But AG=AC+CG and AF=AB+BF, so AC=AB. Q.E.D. #### Swammerdami

Staff member
Surely there's a flaw in this proof! Some triangles are NOT isoceles, right?

It's a good challenge to find the flaw. (I think this "proof" is much more interesting, and challenging, than the usual 1=2 proofs which divide by zero, or equate positive and negative square roots.)

#### lpetrich

##### Contributor
I used Mathematica and analytic geometry, finding the coordinates of the points. I also looked online.

The solution: G is between A and C, and F is outside of A and B in the B direction:
AC = AG + CG
AB = AF - BF

I verified it with my Mathematica calculation:

A = {0,0}, B = {1,0}, C = (1+2a, (1+2a)*(2t)/(1-t2)} where a is arbitrary and t = tan((BAC)/2)
AB = 1, AC = (1+2a)*(1+t2)/(1-t2)
BD = CD = (1/2)*BC = sqrt(t2 + 4t2*a + (1+t2)2*a2)/(1-t2)

D = (1+a, (1+2a)*(t)/(1-t2)}
E = (1 + a*(1+t2))/(1 - t2) * {1,t}
AE = (1+(1+t2)*a)*(1+t2)/(1-t2)
BE = CE = sqrt(1+t2)*sqrt(t2 + 4t2*a + (1+t2)2*a2)/(1-t2)

F = (1+(1+t2)*a)/(1-t2) * {1,0}
G = (1+(1+t2)*a)/(1-t2) * {1-t2,2t}/(1+t2
AF = AG = (1+(1+t2)*a)/(1-t2)

EF = EG = (1+(1+t2)*a) * t/(1-t2)
BF = CG = (t2+(1+t2)*a) /(1-t2)

AB = AF - BF
AC = AG + CG

Here, |t| < 1 and a > 0, so that the distance values come out positive, so they don't have to have reversed signs to be absolute values.

#### lpetrich

##### Contributor
Here are some more problems:

For integers a, b, c, d:

a*b - 2*c*d = 3
a*c + b*d = 1

For nonnegative integers a, b, m, n:

5m + 5n = a2 + b2

For positive integers a, b, c, n:

a*b*c - 1 = n * (a-1) * (b-1) * (c-1)

#### Swammerdami

Staff member
AB = AF - BF
AC = AG + CG
Yes, an easier way to see this is to draw the diagram carefully, making the angle bisector a REAL bisector, and so on.

Another way — perhaps quite difficult — is to examine the proof carefully and see that EVERY step is correctly reasoned EXCEPT the final, implicit assumption that B is between A and F and C is between A and G.

If I'd realized someone might approach the problem the way lpetrich did, I'd have turned myself into the police for Excessive Sadism. Here are some more problems:
...

For nonnegative integers a, b, m, n:

5m + 5n = a2 + b2
m = n = 0
a = b = 1

For positive integers a, b, c, n:

a*b*c - 1 = n * (a-1) * (b-1) * (c-1)
a = b = c = n = 1

Did you want ALL the solutions? #### lpetrich

##### Contributor
Yes all solutions to that one.

a*b - 2*c*d = 3
a*c + b*d = 1

I note that at least one of a, b, c, d must be nonzero. Also that if a, b, c, d is a solution, then -a, -b, -c, -d is also a solution. Since at least one of them is a solution, the solutions are in reverse-sign pairs.

First, multiply the two equations by b and 2c then add them. In short, (b,2c). That gives us
a*(b2+2c2) = 3b + 2c
Repeating for (-c,b) gives us
d*(b2+2c2) = - 3c + b
For (a,2d),
b*(a2+2d2) = 3a + 2d
For (-d,a),
c*(a2+2d2) = -3d + a

Multiplying the a equation by (a2+2d2) and substituting the b and c equations into the rhs of the a equation gives us
a*(a2+2d2)*(b2+2c2) = 3*(3a+2d) + 2*(a-3d) = 11*a
One gets the same sort of equation for b, c, and d, and since at least one of them is nonzero, one finds
(a2+2d2)*(b2+2c2) = 11

Since 11 is a prime number, it has factors 11 and 1. So one of the lhs factors must equal 11 and the other one must equal 1. Let
a2+2d2 = 11
b2+2c2 = 1
with a related solution from the reverse assignment: a <-> b and c <-> d.

The second one has solutions b = +- 1 and c = 0, and the first one solutions a = +- 3 and b = +- 1

Setting b = 1, the equations give us a = 3 and d = 1. So we have solutions (a,b,c,d) =
(3,1,0,1)
(1,3,1,0)
(-3,-1,0,-1)
(-1,-3,-1,0)

Michael Penn has an ingenious approach: multiply the equations by (1,sqrt(-2)):
(a + sqrt(-2)*d) * (b + sqrt(-2)*c) = 3 + sqrt(-2)
and likewise for -sqrt(-2).

But factoring over Z(sqrt(-2)) is more difficult than factoring over Z. But I will try.

Let a*b = 3 + sqrt(-2)
Take the complex conjugate:
cjg(a)*cjg(b) = 3 - sqrt(-2)
Multiply them:
|a|2 * |b|2 = 11
and thus one of |a|2 and |b|2 must equal 1 and the other must equal 11. Thus we get one of them equaling 1 and the other equaling 3 + sqrt(-2).

#### Swammerdami

Staff member
Here are some more problems:

For integers a, b, c, d:

a*b - 2*c*d = 3
a*c + b*d = 1

For nonnegative integers a, b, m, n:

5m + 5n = a2 + b2

For positive integers a, b, c, n:

a*b*c - 1 = n * (a-1) * (b-1) * (c-1)

For the 1st problem
a = 1; b = 3; c = 1; d = 0
(also the negatives of these numbers)

The 2nd problem is the only one for which my solution method was elegant.
We will seek only existence for given m, n. "Congruent" will mean congruent modulo 4.

There is a theorem (due to whom?) that a positive integer N is the sum of two squares, if and only if every prime divisor p≡3 mod 4 of N occurs with even exponent. In our case N = 5m + 5n. Wlog we rewrite this as N = 5m * (1 + 5t) where t is a non-negative integer. Any solution when m = 0 or 1 leads to solutions for 2,4,6,... or 3,5,7,... respectively since we can just multiply a and b by powers of 5. Hence we pursue only m = 0 or m = 1.

1+5t is always congruent to 2 mod 4; after factoring out 2, the result — call it A — is congruent to 1 or 3 according as t is even or odd. (This residue is not changed by the multiplication by 5m.) The number of prime factors congruent to 3 must be even or odd according as t is even or odd. If t is odd there is no solution, by the theorem. It remains only to show that if t is even there always is a solution. Rather than showing that the primes congruent to 3 occur an even number of times, we show more strongly that they occur zero times. In fact we seek the stronger result that multiples of any (x=4a+3) are never congruent to 1 modulo 25.

I do not know the elegant demonstration of this fact, but it can be confirmed by considering 25 cases. Unless I've made an error, I have done so.

This leads to an infinite set of solutions to lpetrich's 2nd question. The 2nd simplest is
5^1 + 5^1 = 1^2 + 3^2
and the 3rd simplest is
5^0 + 5^2 = 1^2 + 5^2
For the 3rd problem
1*1*1 - 1 = N * 0*0*0
2*2*2 - 1 = 7 * 1*1*1
2*2*4 - 1 = 5 * 1*1*3
2*4*8 - 1 = 3 * 1*3*7
3*5*15 - 1 = 2 * 2*4*14

#### lpetrich

##### Contributor
For 5m + 5n = a2 + b2

There are trivial solutions: a = 5m/2, b = 5n/2 for m, n even, and m <-> n.

Take modulo 8. Even m, n give 1, and odd m, n give 5. Squares of 0 to 7 are 0, 1, 4, 1, 0, 1, 4, 1.

For m,n both even or both odd, their sum is 2, and for one even and oen odd, their sum is 6. So a and b must both be odd, and m,n either both even or both odd.

For mod 2 (even/odd), a,b are either both even or both odd, and for mod 4, a,b are both odd.

If a solution exists with both m and n > 0, then the rhs must be divisible by 5. To see how that can happen, take
a2 + b2 = (a + b*i) * (a - b*i)

Also note that 5 = (2 + i) * (2 - i).

Then for every solution (a + b*i) there exists a solution (c + d*i) such that (a + b*i) = (2 + i) * (c + d*i) or
a = 2c - d, b = c + 2d.
c = (2a + b)/5, d = (-a + 2b)/5

But if 5 is not split up, then we have (a + b*i) * (a - b*i) = 5 * (x + y*i) * (x - y*i) and the 5 must join either the (x + y*i) or the (x - y*i), which is not possible. Thus, the 5 must be split up into (2 + i) * (2 - i) or (1 + 2i) * (1 - 2i) or (-1) * each factor.

Thus, for m, n > 0, we find
5m-1 + 5n-1 = c2 + d2

This process can be continued until we find
52n + 1 = a2 + b2

for n getting n - m and a, b getting those transforms.

Complex integers are a Unique factorization domain meaning that any factorization of a complex integer into complex integers is unique to within multiplication by units. Those are anything with a power that makes 1. For ordinary integers, 1, -1, and for complex integers, 1, i, -1, -i.

Along with complex integers Z(i), Z(sqrt(-2)) is also a UFD, but Z(sqrt(-5)) isn't, because 6 = 2*3 = (1+sqrt(-5))*(1-sqrt(-5))

#### lpetrich

##### Contributor
For the reduced equation, I find nontrivial solutions starting with n = 3: (49,115) alongside (1,125) I have found them as far as I can check.

By taking modulo 5, one can find which solutions can possibly exist. 0 to 4 squared mod 5 is 0,1,4,4,1. Since n = 0 gives only the trivial solution, we consider n > 0. That gives sum 1 mod 5, and that means that one of a,b is 0 mod 5 and the other is 1 or 4 mod 5.

Likewise, considering mod 25, one finds that one of a,b is 0 mod 5 and the other one is 1 or 24 mod 25: 25*k +- 1.

I'll now take on mod 4 for sums of squares a2 + b2 = n.

(even)2 = 0 mod 4 and (odd)2 = 1 mod 4

This means that the sum is 0, 1, or 2.

For 0 mod 4, both a and b are even, and one can take a' = a/2, b' = b/2, n' = n/4.

For 2 mod 4, n is divisible by 2 and not 4, both a and b are odd, and a' = (a+b)/2, b' = (a-b)/2, n' = n/2.

For 1 mod 4, n is odd, one of a is even, and the other one is odd.

Likewise for prime factors more generally, if n is divisible by p2, then a' = a/p, b' = b/p, and n' = n/p2

So we have a reduced equation, a2 + b2 = n, where each prime factor of n appears exactly once.

If a prime factor can be expressed as the sum of two squares, then it has the form p = (p1 + i*p2) * (p1 - i*p2). Expressing a2 + b2 = (a + i*b) * (a - i*b), then (a + i*b) must be divisible by (p1 + i*p2) or (p1 - i*p2).

That means that one can divide out every prime factor that is the sum of two squares.

So one is left with a2 + b2 = n where the prime factors of n cannot be expressed as sums of squares, and appear exactly once.

I'm stumped on how to proceed further. I did a numerical search, and I could not find any such n that was the sum of two squares. I also found that every prime up to 7901 that is 1 mod 4 is the sum of two squares, and uniquely so. That is not a proof for all such primes, of course, but I don't know how to prove that.

#### lpetrich

##### Contributor
If a number has factors that are sums of two squares, then that number itself is the sum of two squares: the Diophantus- Brahmagupta–Fibonacci identity
(a2 + b2) * (c2 + d2) = (a*c - b*d)2 + (a*d + b*c)2 = (a*c + b*d)2 + (a*d - b*c)2 Fermat's theorem on sums of two squares - every prime that is 1 mod 4 can be expressed as the sum of two squares. It can also be proved that this decomposition is unique.

#### lpetrich

##### Contributor
I'll now take on
(a*b*c - 1) = n*(a-1)*(b-1)*(c-1)
where a, b, c, n are positive integers with a, b, c > 1.

It is easy to prove that n decreases as a, b, c increase. Take (a*x-1)/(a-1) and d/da: - (x-1)/(a-1)2 -- negative.

So if b >= a and c >= a, then n <= n(b = c = a).

n = 1 is only possible for a, b, c -> infinity, so we consider n = 2 to get upper bounds on the smallest of a, b, c.

n = 2: a <= 4
n = 3: a <= 3
n >= 4: a <= 2

n = 2, a = 2: no solutions
n = 2, a = 3: one solution: b = 5, c = 15
n = 3, a = 2: one solution: b = 4, c = 8
n = 3, a = 3: no solutions
n = 4, a = 2: no solutions
n = 5, a = 2: one solution: b = 2, c = 4
n = 6, a = 2: no solutions
n = 7, a = 2: one solution: b = 2, c = 2

Thus, all the solutions (a,b,c,n):
(2,2,2,7)
(2,2,4,5)
(2,4,8,3)
(3,5,15,2)

Now a simpler problem:
(a*b - 1) = n*(a-1)*(b-1)

For a = b = 2, n = 3, so one need only try n = 2. For a = b, one finds a = 3, so a = b = 3, n = 2 is the only other solution. Thus,
(2,2,3)
(3,3,2)

#### Swammerdami

Staff member
If a number has factors that are sums of two squares, then that number itself is the sum of two squares: the Diophantus- Brahmagupta–Fibonacci identity
(a2 + b2) * (c2 + d2) = (a*c - b*d)2 + (a*d + b*c)2 = (a*c + b*d)2 + (a*d - b*c)2 Fermat's theorem on sums of two squares - every prime that is 1 mod 4 can be expressed as the sum of two squares. It can also be proved that this decomposition is unique.
These two theorems are well-known and immediately prove the IF part of the Theorem I needed for the solution in a Spoiler above. How about the ONLY IF part?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The second theorem lpetrich mentions is called Fermat's Christmas Theorem; it is very famous and has several proofs. Two of the proofs have special interest.

(a) One proof is contained in a journal paper touted as "the shortest math paper" ever! (Or the "single-sentence proof.") The proof simply displays a 1:1 mapping between two classes defined by math expressions. Returning to the theme of the thread ("YouTube"), Mathologer apparently posted a video about that "shortest paper."

(b) In the comments to Mathologer's YouTube someone known only by nickname (since then perhaps he's come forth) showed how to map that proof to a combinatorial argument based on counting windmill shapes!

So Mathologer prepared this video explaining the windmill counting scheme. It's breathtakingly beautiful. The windmill shapes are counted in two completely different ways; they aren't really counted: we just want to know if the total count is even or odd. The different ways involve two different ways of arranging the shapes into pairs; if one is left over you know the number is odd.

Both the AT-LEAST-ONE and the AT-MOST-ONE parts of Fermat's Christmas Theorem are non-trivial. The windmill counting only does the AT-LEAST-ONE. (Mathologer's YouTube finishes with a demonstration of AT-MOST-ONE, which he does with a long series of simple algebraic manipulations.)

(ETA: I clicked on the video just now. If you're in a hurry to get to the windmill counting proof you might want to skip the first 11 minutes or more. OTOH the whole talk may be fun. For example, a survey of math teachers ranked the Christmas Theorem #10 on a "Most Beautiful Theorems" list! ... And that's even without this gorgeous windmill-counting proof.)

Last edited:

#### Swammerdami

Staff member
Here's a fun little puzzle (though probably not on YouTube). An interesting comment about difficulty can be applied to the solution; extra credit if you guess what I'm talking about!

Given four distinct points in the plane, there are six pairwise distances among those points.
You are told that there are only two distinct values among those six distances. Show us all such solutions.

#### lpetrich

##### Contributor
Another Michael Penn one, from Srinivasa Ramanujan:

$$\textstyle{ \lfloor \frac{n}{3} \rfloor + \lfloor \frac{n+2}{6} \rfloor + \lfloor \frac{n+4}{6} \rfloor = \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n+3}{6} \rfloor }$$

The |_ and _| mean "floor", the largest integer <= its arg.

What integer n values satisfy this equation?

#### lpetrich

##### Contributor
Turning to Swammerdami's problem, here are all the possible sets of point distances, to within point relabelings:
1. 12 / 13 14 23 24 34
2. 12 13 / 14 23 24 34
3. 12 34 / 13 14 23 24
4. 12 13 14 / 23 24 34
5. 12 13 24 / 14 23 34
I checked that these are the only possible splits using Mathematica, but it should be easy with a program in (say) Python. Some of its standard library is convenient for that, like a permutation generator in itertools and the set() function for unique-member lists

For the first one, triangles 134 and 234 are both equilateral triangles, and they share side 34. Segment 12 thus has length sqrt(3) relative to the others.

For the second one, 234 is an equilateral triangle, and point 1 a little beyond the center of side 23. 12 and 13 have length (sqrt(3)-1)/sqrt(2) times the lengths of the others.

For the third one, 1324 is a square, and 12 and 34 are its diagonals, with length sqrt(2) relative to the others.

For the fourth one, 234 is an equilateral triangle, and 1 is its center, That makes 12 have length 1/sqrt(3) times 23's length.

For the fifth one, the points are in two sequences with equal-length segments: 3124 and 1432. One set of segments will be longer than the other set, with a length ratio (sqrt(5)+1)/2 or the reciprocal of (sqrt(5)-1)/2.

#### Swammerdami

Staff member
Another Michael Penn one, from Srinivasa Ramanujan:

$$\textstyle{ \lfloor \frac{n}{3} \rfloor + \lfloor \frac{n+2}{6} \rfloor + \lfloor \frac{n+4}{6} \rfloor = \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n+3}{6} \rfloor }$$

The |_ and _| mean "floor", the largest integer <= its arg.

What integer n values satisfy this equation?
Is this a trick question? I think it's true for ALL integers. This is trivially true when n is a multiple of 6. Step through the other five cases by adding 1 (none of the expressions changes value), then 2 (one increments on the left, another on the right), then 3 (ditto), 4 (ditto), 5 (no change).

Turning to Swammerdami's problem, here are all the possible sets of point distances, to within point relabelings:
...
I checked that these are the only possible splits using Mathematica, but it should be easy with a program in (say) Python. Some of its standard library is convenient for that, like a permutation generator in itertools and the set() function for unique-member lists
The list you defined may be exhaustive, but you've still missed a solution. (I won't say more, and anyway I always prefer solutions in spoilers ... though there's a fair chance lpetrich will be the only contestant. )

Sometimes it's FUN to use Mathematica or Python as a solving tool, but figuring by hand with paper and pencil is also fun and in some cases (including this one) may be much EASIER.

IIUC, the list you developed via Mathematica was a trivial thing that could have been done readily by hand. AND developing that list was an obstacle anyway; did I mention that you missed one of the SIX solutions?

I think at least five of the six solutions can be found readily by hand. Doing so should bring some JOY. ... And help toward the Extra Credit problem ("An interesting comment about difficulty can be applied to the solution").

#### lpetrich

##### Contributor
I used Mathematica because I wasn't sure of my reasoning. I'll go through my reasoning. For points 1, 2, 3, 4: segments 12, 13, 14, 23, 24, 34.

One segment: 12

Two segments: connected to one of 1, 2 or connected to none of them:
12 13
12 23

Three segments: connected to 1 of the first one or 2 of the second one, making the segments radiate out from one point. Or else connected to 2 or 3 of the first one or 1 or 3 of the second one, making a strip. Or else connecting the two that show up only once:
12 13 14 -- radiating from 1
12 13 24 -- strip 4213
12 13 23 -- triangle 123

Let us look at the remaining segments for each one:
12 13 14 / 23 24 34 -- radiation from 1 + triangle
12 13 24 / 14 23 34 -- two strips: 4213, 1432
12 13 23 / 14 24 34 -- triangle + radiation from 4

So the sixth solution is a version of one of the five ones.

#### Swammerdami

Staff member
I used Mathematica because I wasn't sure of my reasoning. I'll go through my reasoning. For points 1, 2, 3, 4: segments 12, 13, 14, 23, 24, 34.
Mathematica would be useful for a similar but slightly more complicated puzzle. But for this one, the (3;3) partition is most "difficult" but is readily understood: if there's no equilateral triangle then 3 of the distances must be three sides of a quadrilateral, and the remaining three distances will ALSO be three such sides.
So the sixth solution is a version of one of the five ones.
Obviously. The punchline to the puzzle depends on which of the six solutions you find most difficult to find. You might consider the final solution you find to be most difficult, while the "punchline" depends on the hardest-to-find being one you've already found. (Guess which one I mean?)

Two related puzzles:
(a) FIVE points in the plane, with only two distinct point-to-point distances.
(b) Five points in the plane, with only three distinct point-to-point distances.

I am NOT masochistic enough to attempt (b). I might add it to my already-overlong To-Do list. ETA: In fact, unless you have a VERY clever program to let a machine do the work, (b) might be ridiculously cumbersome. In an effort to make amends I've crossed it out.

#### lpetrich

##### Contributor
That is correct about the Ramanujan floor puzzle. Set n = 6*m + k where m is an integer and k is an integer between 0 and 5 inclusive.

Then, floor((n+a)/b) = (6/a)*m + floor((k+a)/b).
For m, 6/3 + 6/6 + 6/6 = 4, 6/3 + 6/6 = 4
For k,
0: 0+0+0 = 0, 0+0 = 0
1: 0+0+0 = 0, 0+0 = 0
2: 0+0+1 = 1, 1+0 = 1
3: 1+0+1 = 2, 1+1 = 2
4: 1+1+1 = 3, 2+1 = 3
5: 1+1+1 = 3, 2+1 = 3

#### lpetrich

##### Contributor
Since there are 5 points, there are 5*(5-1)/2 = 10 line segments connecting them. That means that we must consider sets of 1, 2, 3, 4, and 5 segments.

I used Mathematica to sort out the possibilities. Here are all the distinct network topologies for each number of segments and number of segments in its complement (total segments - those segments). When both a set of segments and its complement have the same number, then we have (number of all that have distinct flipped versions, number of all with the same flipped version)
• 2: 1
• 0 - 1: 1
• 3: 3
• 0 - 3: 1
• 1 - 2: 1
• 4: 6
• 0 - 6: 1
• 1 - 5: 1
• 2 - 4: 2
• 3 - 3: 3 (1,1)
• 5: 10
• 0 - 10: 1
• 1 - 9: 1
• 2 - 8: 2
• 3 - 7: 4
• 4 - 6: 6
• 5 - 5: 6 (2,2)
For 5 points, many of the topologies can easily be ruled out because of the impossibility of creating equilateral triangles with distinct vertices that satisfy that. That rules out all those that contain squares with both diagonals and squares with center points connected to 3 or 4 vertices. Those are all of (0,10), (1,9), and (2,8), and this criterion leaves 1 of (3,7), 4 of (4,6), and all of (5,5).

Of these, all but one of them can be ruled out by taking the complement segments and finding the possible lengths of the original segments.

The remaining one is a pentagonal topology for both the original segments and their complement.

This leads to the only solution: a regular pentagon and a pentagram.

#### lpetrich

##### Contributor
Instead of 5 points with 3 different lengths, I will try 4 points with 3 different lengths. The 3-point case is trivial.
Let us first see how the edges divide up:
• 1 - 1 - 4
• 1 - 2 - 3
• 2 - 2 - 2
There are solutions for all three cases, solutions with one free parameter.

I've considered 6 points. For two segment lengths, there are no solutions. For three, there is at least one: a regular hexagon with a hexagram and the three diagonals.

For an n-gon, this kind of solution will always exist, and will contain floor(n/2) different segment lengths.

#### Swammerdami

Staff member
Still waiting for someone to list the six solutions to the (4,2) problem — 4 points in the plane which have only 2 unique distances — and explain what's "paradoxical" about the difficulty of finding the (allegedly) most difficult-to-find of the six.

Meanwhile ...

...
Two related puzzles:
(a) FIVE points in the plane, with only two distinct point-to-point distances.
(b) Five points in the plane, with only three distinct point-to-point distances.

I deleted the suggestion about the (5,3) investigation for fear of being arraigned by the Society for the Prevention of Cruelty by Sadists. However masochism is not a crime, and I have pursued the problem a bit.

I will list some of the (5,3) solutions. The list will not be exhaustive. Would it be a fun contest to see who can find more solutions?

I wrote a program with the same goal as lpetrich's: to find all classes of distance assignments. There are 142 classes, but 108 of those derive simply: start with one of the six (4,2) solutions, and add a 5th point. I'll refer to those 108 classes as "derived classes."

This leaves 34 classes for which any 4-sized subset uses three distances. I'll refer to these as "non-derived classes."

Note that a class, whether "derived" or "non-derived," may or may not lead to a satisfactory geometry, and might lead to two or more such geometries.

From the square (a (4,2) solution with distances 1 and 1.414), a 5th point can be added in five places, all along a lateral bisector. An isosceles triangle can be added to a side, either pointing in or pointing out, and with side either 1 or 1.414. This is four solutions; a fifth solution is to place the 5th point in the center of the square. Is that it for the square?

Another (4,2) solution is the four-fifths of a pentagon with distances 1 and 1.618. Now there seem to be eight derived solutions: add an isosceles triangle to either of the two parallel sides, pointing either out or in, and with side either 1 or 1.618. However these eight solutions are just seven because the (5,2) solution -- the perfect 5-star -- is derived with EITHER an in-pointing 1.618 isosceles triangle from the short side OR an out-pointing 1.0 isosceles triangle from the long side.

The other four (4,2) solutions are described by an equilateral triangle and a 4th point on its bisector. For these cases I've given up (temporarily?) after deriving just one (5,3) solution: Add a 4th point in the center of the triangle and a 5th point that forms a triangle congruent to the little triangles just formed.

So far we're up to 13 solutions total. Now let's look at "non-derived" solutions.

I've listed the 34 non-derived cases in a Spoiler; I describe the notation with this example (first in the list):
. . . . . . . . AAAA // ABC // BA // C . #Equ= 2 #Iso= 6
Reading left to right, point #5 has distance A to points #1,#2,#3,#4; point #4 has distances A,B,C to points #1,#2,#3 respectively; point #3 has distances B,A to points #1,#2 and the distance from #2 to #1 is C. (The software reports, redundantly, that the case has 2 equilateral triangles and 6 isosceles triangles. (Equilaterals are included in the Isosceles count.)

If you think about this with a pencil, you find one solution: a 1-by-1.732 rectangle with its center. (1.732 is the square root of 3.)

Here's the 2nd non-derived class
. . . . . . . . AAAA // ABC // BB // C . #Equ= 1 #Iso= 8
Again, the AAAA to start means point #5 is the center of a circle, with the other 4 points somewhere on the circumference. Think of angles rather than distances; the central angle (call it t) for points 1-3 equals the angle for 3-2 which equals the angle for 2-4; and the 4-1 angle is 60 degrees. This means t is either 100 degrees or 140 degrees; either leads to solution.

I've not looked at the other 32 non-derived cases. Any other masochists out there? AAAA // ABC // BA // C . #Equ= 2 #Iso= 6
AAAA // ABC // BB // C . #Equ= 1 #Iso= 8
AAAB // AAC // AB // C . #Equ= 1 #Iso= 7
AAAB // AAC // BB // C . #Equ= 0 #Iso= 7
AAAB // AAC // BC // C . #Equ= 0 #Iso= 8
AAAB // ABC // AB // C . #Equ= 1 #Iso= 7
AAAB // ABC // BA // C . #Equ= 1 #Iso= 5
AAAB // ABC // BB // C . #Equ= 0 #Iso= 7
AAAB // ABC // BC // C . #Equ= 0 #Iso= 7
AAAB // ABC // CB // C . #Equ= 0 #Iso= 8
AAAB // ACC // AC // B . #Equ= 2 #Iso= 6
AAAB // ACC // BA // C . #Equ= 1 #Iso= 6
AAAB // ACC // BB // C . #Equ= 0 #Iso= 7
AAAB // ACC // BC // B . #Equ= 1 #Iso= 6
AAAB // ACC // BC // C . #Equ= 1 #Iso= 7
AAAB // ACC // CB // C . #Equ= 0 #Iso= 8
AAAB // BCC // AB // C . #Equ= 1 #Iso= 6
AAAB // BCC // AC // B . #Equ= 2 #Iso= 6
AAAB // BCC // BA // C . #Equ= 1 #Iso= 7
AAAB // CCC // AB // C . #Equ= 2 #Iso= 6
AABB // AAB // CC // C . #Equ= 2 #Iso= 6
AABB // AAC // BC // C . #Equ= 0 #Iso= 8
AABB // ABA // CC // C . #Equ= 1 #Iso= 6
AABB // ABC // BA // C . #Equ= 0 #Iso= 6
AABB // ABC // BC // C . #Equ= 0 #Iso= 7
AABB // ACA // CA // B . #Equ= 0 #Iso= 6
AABB // ACA // CB // B . #Equ= 0 #Iso= 6
AABB // ACA // CC // B . #Equ= 0 #Iso= 6
AABB // ACB // CA // B . #Equ= 1 #Iso= 4
AABB // ACB // CC // B . #Equ= 1 #Iso= 5
AABB // ACC // CA // B . #Equ= 0 #Iso= 6
AABB // CCA // CC // B . #Equ= 0 #Iso= 6
AABC // ABA // BC // C . #Equ= 0 #Iso= 5
AABC // BBA // CC // A . #Equ= 1 #Iso= 3

Staff member

#### lpetrich

##### Contributor
12 points on a 7-pointed star shape (heptagram), making both 4 equilateral triangles and 3 squares. The heptagram has each vertex connected to those nearest the opposite side from it.

The first step is constructing that heptagram. For that purpose, we need to consider constructing a regular polygon. For a regular n-gon, the construction involves trigonometric functions of the angle (pi/n) or (180d/n).

That may seem very difficult, but trigonometric identities yield polynomial equations. Like cos(2x) = cos(x)2 - sin(x)2 and sin(2x) = 2*cos(x)*sin(x) along with cos(x)2 + sin(x)2 = 1.

In general, for nonnegative integer n, cos(n*x) = T(n,cos(x)) and sin(n*x) = U(n-1,cos(x))*sin(x) where T and U are the Chebyshev polynomials of the first and second kinds. These definitions give a variety of features of them -- recurrence relations, differential equations, special values, generating functions, etc. -- including having integer coefficients.

So T(n,cos(pi/n)) = cos(pi) = -1, so one has a polynomial equation to solve for cos(pi/n). At first sight it seems very difficult to do this solution, but there are a variety of simplifications that one can use.

Before continuing, note that this equation's solutions are not unique. All the solutions can be expressed as cos((2k+1)*pi/n), for k = 1 to (n-1), though that means that if one has cos(pi/n), one can find all the others using Chebyshev polynomials of it. Furthermore, replacing k with (n-k-1) gives the same solution, and if n is odd, then k = (n-1)/2 gives -1. This means that for n even, this becomes the square of an equation with degree (n/2), and for n odd, the square of an equation with degree ((n-1)/2) multiplied by (variable + 1). So the effective degree of this polynomial equation is reduced by a factor of 2.

For n even, one can use the half-angle formulas to get cos(pi/(2n)) in terms of cos(pi/n) -- x becomes sqrt((x+1)/2). Likewise, for n a multiple of 3, one can do something similar with a cubic equation, though that is much more complicated.

So let us consider n odd. This degree reduction explains why cos(pi/3) = 1/2 with no square root and cos(pi/5) = (sqrt(5)+1)/4 and not some more complicated equation. Because (3-1)/2 = 1 and (5-1)/2.

For a heptagon, one thus has a cubic equation to solve, and that is very complicated.