<s>[INST] Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square Find all the positive integers $k$ for which $7 \times 2^k+1$ is a perfect square.
The only value of $k$ I can find is $5$. I am not sure how to find every single one or the proof, I simply used trial and error.
 [/INST] Since $7\cdot 2^k+1$ is odd we have that
$$7\cdot 2^k+1=(2n+1)^2=4n^2+4n+1 \iff7\cdot 2^k=2^2n(n+1) \iff 7\cdot 2^{k-2}=n(n+1)$$
and since one among $n$ and $n+1$ must be odd therefore amd since $7$ is prime,  one of them must be equal to $7$ and then the unique solution is $2^{k-2}=8 \implies k=5$.
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