3^x + 4^y equals 5^z has Unique Solution

Theorem
The Diophantine equation:
 * $3^x + 4^y = 5^z$

has exactly one solution in (strictly) positive integers:


 * $3^2 + 4^2 = 5^2$

Proof
Rewriting our Diophantine equation Modulo 4 we have:
 * $\paren {-1}^x + 0^y \equiv 1^z \pmod 4$

Therefore, x is even

Rewriting our Diophantine equation Modulo 3 we have:
 * $0^x + 1^y \equiv \paren {-1}^z \pmod 3$

Therefore, z is even

Since x and z must both be even, we will rewrite x as $2$r, z as $2$s and notice that $4$ is $2^2$

We now have:

We can see that the is a product consisting entirely of $2$s, therefore:


 * $\paren{5^s + 3^r} = 2^u$ and
 * $\paren{5^s - 3^r} = 2^v$

Where $u \gt v$ and $u + v = 2y$

Solving for $5^s$, we add the two equations and get:

Solving for $3^r$, we subtract the two equations and get:

Since both $5^s$ and $3^r$ are odd, $v$ must be equal to $1$.

Let $t = u - v$

We now have:
 * $5^s = \paren {2^t + 1}$
 * $3^r = \paren {2^t - 1}$

Let us now consider the second equation Modulo 3:
 * $0 \equiv \paren{-1}^t - 1 \pmod 3$

Therefore t must be even.

We will rewrite the second equation above with t as $2$w

We can see that the is a product consisting entirely of $3$s, therefore:


 * $ 3^m = \paren {2^w + 1}$ and
 * $ 3^n = \paren {2^w - 1}$

Where $m \gt n$ and $m + n = r$

Subtracting the $2$ equations above we get:

Therefore $m = 1$ and $n = 0$

From above
 * $ 3^m = \paren {2^w + 1}$

Therefore $w = 1$

Recall that $t = 2w$

Therefore $t = 2$

And:
 * $5^s = \paren {2^t + 1}$
 * $3^r = \paren {2^t - 1}$

Therefore $r = 1$ and $s = 1$

Finally recall that $x = 2r$ and $z = 2s$

Therefore we arrive at the only possible solution in (strictly) positive integers : $x = 2$, $y = 2$ and $z = 2$