YouTube Math-Puzzle and Math-Instruction Channels

[lpetrich]

These YouTube channels have lots of videos showing solutions of mathematics puzzles:

• Michael Penn - YouTube
• MindYourDecisions - YouTube
• letsthinkcritically - YouTube
• Mind your BRAIN - YouTube
• PreMath - YouTube

These are more about math exposition and instruction:

• Numberphile - YouTube
• Mathologer - YouTube
• 3Blue1Brown - YouTube

Some of the puzzles are problems from various Mathematics Olympiads:

• Indian Mathematical Olympiad | 1992 Question 8 - YouTube -- \(2^x + 3^y = n^2\)
• Equation on Pythagorean Triples? | IMO Shortlist 1991 - YouTube -- \(3^x + 4^y = 5^z\)
• Australian mathematical Olympiad - YouTube -- \(2^x - 3^y = 7\)

Note: all the problem variables are nonnegative integers.

 

[lpetrich]

I'll do all three math puzzles.

---

2^x + 3^y = n²

Do modular arithmetic, modulo 4:

x: 2^x -> 0: 1, 1: 2, (2: 0)
y: 3^y -> (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 - 3^2y' = (n - 3^y')(n + 3^y')
No solution.

For x = 1, try mod 8:
y: 3^y -> (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:
2^x = n² - 3^2y' = (n - 3^y')(n + 3^y')

---

3^x + 4^y = 5^z

---

2^x - 3^y = 7

Solution:

 

[lpetrich]

2^x + 3^y = n²

Do modular arithmetic, modulo 4:

x: 2^x -> 0: 1, 1: 2, (2: 0)
y: 3^y -> (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 - 3^2y' = (n - 3^y')(n + 3^y')
No solution.

For x = 1, try mod 8:
y: 3^y -> (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:
2^x = n² - 3^2y' = (n - 3^y')(n + 3^y')

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

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

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]

3^x + 4^y = 5^z

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

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

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

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

2^2y = 5^2z' - 3^2x' = (5^z' - 3^x') (5^z' + 3^x')

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

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

Taking mod 8, 3^x' 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]

2^x - 3^y = 7

First, since 2^x > 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 2^x = 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

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

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

2⁴ - 3² = 16 - 9 = 7

 

[lpetrich]

More:

• SyberMath - YouTube
• PreMath - YouTube

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

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

 

[none]

42

 

[lpetrich]

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

2^a + 3^b + 5^c = 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]

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) x^a - y^b = 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]

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

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

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

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]

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 2^a + 3^b + 5^c = 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
\( (x - 2)(2x - 1)(7 x^2 + 5 x + 7) \)

One can use the quadratic formula on the quadratic factor, giving
\( \displaystyle{ x = \frac{ - 5 \pm 3 i \sqrt{19} } {14} } \)

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

 

[lpetrich]

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 \( \sqrt[7]{x} - \sqrt[5]{x} = \sqrt[3]{x} - \sqrt{x} \) ?

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]

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

---

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

 

[lpetrich]

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 \( x = 3^{2/3} 5^{1/3} \)

---

\( \sqrt[7]{x} - \sqrt[5]{x} = \sqrt[3]{x} - \sqrt{x} \)
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:
\( \sqrt[7]{x} - \sqrt[5]{x} = \sqrt[3]{x} - x \)
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.

 

[lpetrich]

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]

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]

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]

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]

From Let's Think Critically: nonnegative integers a, b, n with 2^a + 3^b = 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
\( \displaystyle{ u = C x^{-1/2} + \frac15 x^2 } \)