Berkeley Math Circle Monthly Contest 2

Solution:
We prove this by induction on $n$.

If $n=1,2$, or $3$, then $x=1$ works.

Suppose that $x^{2}-17$ is divisible by $2^{n}$ and $n \geq 3$. We seek to find $y$ such that $y^{2}-17$ is divisible by $2^{n+1}$. Let
$$x^{2}-17=k \cdot 2^{n} \text{.}$$
If $k$ is even, we are done since $x^{2}-17$ is divisible by $2^{n+1}$. If $k$ is odd, $k=2 m+1$, we have
$$\begin{aligned} \left(x+2^{n-1}\right)^{2}-17 & =x^{2}+2 \cdot x \cdot 2^{n-1}+2^{2(n-1)}-17 \\ & =x^{2}-17+x \cdot 2^{n}+2^{2 n-2} \\ & =(2 m+1) \cdot 2^{n}+x \cdot 2^{n}+2^{2 n-2} \\ & =2^{n}(x+1)+2^{n+1}\left(m+2^{n-3}\right) \end{aligned}$$
Since $x$ is obviously odd and $n \geq 3$, this is a multiple of $2^{n+1}$, and $y=x+2^{n-1}$ works.