# Find all $x,y\in \mathbb{N}$ such that: $2^x+17=y^2$.

Find all $x,y\in \mathbb{N}$ such that: $2^x+17=y^2$.

Its easy to find that $x=6$ is the only even value for $x$, the others have to be odd. One more thing is that we get $y^2 \equiv 19 \pmod p$, for every prime factor of $x$. But I have no ides what next to do.

• Just a little useless remark: This is an equation "of Ramanujan-Nagell type": there are some interesting properties of this kind of equations on en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation; but it really doesn't help your specific problem :-) – PseudoNeo Sep 8 '15 at 18:22
• – Zev Chonoles Sep 8 '15 at 18:23
• @ZevChonoles It is a solution to the problem, but I think there shouldnt be a more elementar way to do it. – HeatTheIce Sep 8 '15 at 18:33
• What is the source of this problem? – user236182 Nov 30 '15 at 18:18
• Why are you claiming that $2^x\equiv 2\pmod{p}$ for every prime factor $p$ of $x$? – user236182 Nov 30 '15 at 18:48