How to find all solutions to $2^x=y^2+15$? This should be some math Olympiad question. I think the digits in the ones place of $2^x$ are 4 8 6 2 4 so it means $y^2$'s ones place should be 9, 3, 1, 7, 1 where 3 and 7 are impossible so it's 9 and 1.
Is this right? And I just get stuck...
Checking modulo 3 we conclude that $x$ must be even, as $2^x \equiv 2$ (mod 3) for odd $x$ and $y^2 \equiv 1$ (mod 3) for all integers $y$ not a multiple of 3--and $y$ cannot be a a multiple of 3 [why].
Thus:
As $x$ is even, the equation $2^x-y^2 = 15$ implies $(2^{x/2}-y)(2^{x/2}+y) = 15$; $x/2$ an integer, which assuming WLOG that $y$ is positive, implies that $|2^{x/2}+y|$ is a positive integer no larger than $15$ and that in fact divides 15. That leaves few cases, few enough to check exhaustively.
This leaves us with 2 solutions (a) $x=6$ and $y=7$, and (b) $x=4$ and $y=1$.
Assuming you're referring to positive integers, note that you have
$$2^x \equiv y^2 \pmod 5 \tag{1}\label{eq1A}$$
Since $y$ can't be a multiple of $5$, you have $y^2$ being congruent to either $1$ or $4$ modulo $5$. However, this means that $x$ must be even, say $x = 2z$ for some integer $z$. Then you have
$$2^{2z} - y^2 = 15 \implies (2^z - y)(2^z + y) = 15 \tag{2}\label{eq2A}$$
Since $15 = 3 \times 5$, this leaves very few factors to check. In particular, since $0 \lt 2^z - y \lt 2^z + y$, you have
$$2^z - y = 1, 2^z + y = 15 \implies 2y = 14, y = 7, 2^z = 8, z = 3 \tag{3}\label{eq3A}$$
$$2^z - y = 3, 2^z + y = 5 \implies 2y = 2, y = 1, 2^z - 4, z = 2 \tag{4}\label{eq4A}$$
Thus, the only the solutions for $(x,y)$ are $(4,1)$ (i.e., $16 = 1 + 15$) and $(6,7)$ (i.e., $64 = 49 + 15$).
