YouTube Math-Puzzle and Math-Instruction Channels

[lpetrich]

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
\( x^4 - 14 x^2 - x + 42 = (x - 2) (x + 3) (x^2 - x - 7) \)

The quadratic part has solutions
\( \displaystyle{ x = \frac12 \left( 1 \pm \sqrt{29} \right) } \)

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]

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]

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]

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

Adjust the bounds:
\( \displaystyle{ \prod_{k=2}^N (k-1) \prod_{k=4}^{N+2} \frac{1}{k-1} \prod_{k=3}^{N+1} (k^2-k+1) \prod_{k=2}^N \frac{1}{k^2-k+1} } \)

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]

I'll try 2^a + 3^b = 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 2^a + 1 = n² Subtracting 1 gives
2^a = n² - 1 = (n - 1) * (n + 1)

Splitting into factors,
2^a2 = n + 1 and 2^a1 = n - 1
where a1 and a2 are nonnegative integers that add to make a. Some rearrangement gives us
2^a2 - 2^a1 = 2
This means that a2 > a1 and a1 > 0, and
(2^a2-a1 - 1) 2^a1-1 = 1
giving a1 = 1 and
2^a2-a1 -1 = 1 then 2^a2-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, 2^2a' + 3^b = n² giving
3^b = n² - 2^2a' = (n + 2^a') * (n - 2^a')

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

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

2^2a''+1 = 3^2b' - 1 = (3^b' - 1) * (3^b' + 1)
Splitting the power of 2 as 2a''+1 = a1 + a2,
2^a2 = 3^b' + 1 and 2^a1 = 3^b' - 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]

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]

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]

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]

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)}
AD = sqrt((1-t²+t⁴)+2(1+t⁴)*a + (1+t²)²*a²)/(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]

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:

5^m + 5ⁿ = a² + b²

---

For positive integers a, b, c, n:

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

 

[Swammerdami]

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:

5^m + 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]

Yes all solutions to that one.

About my first 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]

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:

5^m + 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

Spoiler

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.

Spoiler

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 = 5^m + 5ⁿ. Wlog we rewrite this as N = 5^m * (1 + 5^t) 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+5^t 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 5^m.) 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

Spoiler

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]

For 5^m + 5ⁿ = a² + b²

There are trivial solutions: a = 5^m/2, b = 5^n/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
5^m-1 + 5^n-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]

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]

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]

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]

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.)

 

[Swammerdami]

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]

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?