<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] If $y=0$, we have $2^x=z^2-1=(z+1)(z-1)$, for which the only solution is $x=z=3$. That is, we have $z+1=2^a$ and $z-1=2^b$ with $a+b=x$. But then $2=(z+1)-(z-1)=2^a-2^b$, for which the only solution is $a=2$ and $b=1$, from which $x=z=3$ follows.
If $y\ge1$, then $2^x$ is a square mod $3$, which implies $x$ is even. Writing $x=2w$, we have
$$3^y=z^2-2^{2w}=(z+2^w)(z-2^w)$$
which implies $2^{w+1}=3^y-1$. (That is, $z+2^w$ and $z-2^w$ must each be a power of $3$ whose exponents' sum is $y$, so their difference, which is $2\cdot2^w$, is the difference of those two powers of $3$. But $2^{w+1}$ is not divisible by $3$, so the smaller exponent must be $0$.) Now if $y$ is odd, then $3^y\equiv3$ mod $8$, which implies $2^{w+1}\equiv2$ mod $8$, so that $w$ can only be $0$, which gives us the solution $(x,y,z)=(0,1,2)$ and no other solutions with $y$ odd. On the other hand, if $y$ is even, say $y=2v$, then, writing $z'=3^v$, we have $2^{w+1}=z'^2-1$, for which we saw in the first paragraph there is only the solution $w+1=z'=3$, giving us the solution $(x,y,z)=(4,2,5)$ and no other solutions with $y$ even.
So that's it: the equation $2^x+3^y=z^2$ has the three solutions $(3,0,3)$, $(0,1,2)$, and $(4,2,5)$ and no others.
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