<s>[INST] Find all natural solutions that satisfy $2^ + 3^ = ^2$ It looks like an easy question but I couldn't find a way to solve it. I found (0,1,2),(3,0,3),(4,2,5) by trial and error and I'm kinda sure they are the only answers but I'm not sure how to prove it.
 [/INST] First consider the cases $x=0$ and $y=0$. If $x=0$, then $3^y=z^2-1=(z-1)(z+1)$ and hence $z-1$ and $z+1$ are two powers of $3$ with distance $2$; that immediately implies $z+1=3, z-1=1$ and so $z=2, y=1$. If $y=0$, then $2^x=z^2-1=(z-1)(z+1)$ and hence $z-1$ and $z+1$ are two powers of $2$ with distance $2$; that immediately implies $z+1=4, z-1=2$ and so $z=3, x=3$. We thus have the solutions $(0,1,2)$ and $(3,0,3)$.
Now let $x,y>0$. Considering the equation mod $3$, we see that the RHS is $\equiv 0,1$ (as it is a square) and the LHS is $\equiv 2^x$. As $2^x\equiv 0\pmod{3}$ is impossible, we must have $2^x\equiv 1\pmod{3}$, which means that $x$ is even. As a consequence, $x=2x'$ for some $x'>0$. Considering the equation mod $4$, we now see that the RHS is $\equiv 0,1$ and the LHS is $\equiv 3^y$. As $3^y\equiv 0\pmod{4}$ is impossible, we must have $3^y\equiv 1\pmod{4}$, which means that $y$ is even. As a consequence, $y=2y'$ for some $y'>0$.
We now have the equation
$$2^{2x'}+3^{2y'}=z^2\Leftrightarrow \left(2^{x'}\right)^2+\left(3^{y'}\right)^2=z^2$$
and so $\left(2^{x'}, 3^{y'}, z\right)$ is a Pythagorean triple; it is even a primitive Pythagorean triple. Such triples are of the form $\left(2uv, u^2-v^2, u^2+v^2\right)$ for coprime integers $u>v$ with distinct parity. We must then have $2uv=2^{x'}$ as both of them are the only even numbers in each triple; this means that $u$ and $v$ are both powers of $2$. However, they must also be coprime and have different parity, so the only option is $u=2, v=1$. This gives $x'=2$ and $u^2-v^2=3$, which is indeed a power of $3$, so $y'=1$. Hence, $x=4$ and $y=2$, which leads to $z=5$. Therefore, we have the additional solution $(4,2,5)$.
In conclusion, the solutions $(0,1,2), (3,0,3), (4,2,5)$ are the only ones.
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