Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square

Question

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.

Answer 1

$$7\times 2^k+1=n^2\iff7\times2^k=(n-1)(n+1)$$

$n$ must be odd so, $7\times 2^{k-2}=(\frac{n-1}{2})(\frac{n+1}{2})$ and $\gcd(\frac{n-1}{2},\frac{n+1}{2})=1$, since $\gcd(\frac{n-1}{2},\frac{n+1}{2})$ divides $\frac{n+1}{2}-\frac{n-1}{2}=1$.

This gives two possibilities:

• $\frac{n-1}{2}=7\Rightarrow n=225\Rightarrow 2^k=32$
• $\frac{n+1}{2}=7\Rightarrow n=169\Rightarrow 2^k=24$

Answer 2

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 $2^{k-2}=8 \implies k=5$.

Answer 3

\begin{align} a^2 &= 7*2^k + 1 \\ \Rightarrow (a^2 - 1) = (a - 1)(a+1) &= 7*2^k \\ \text{For } k > 2: (\frac{a-1 }{2}) (\frac{a+1}{2}) &= 7*2^{k-2} \\ \text{Let } d = (\frac{a-1 }{2}) &\Rightarrow d + 1 = (\frac{a+1}{2}).\\ \Rightarrow d(d+1) &= 7*2^n (n = k - 2\in \mathbb{N})\\ gcd(d,d+1) = 1 &\Rightarrow (d,d+1) = (7,2^n) \text{ or } (2^n,7) \\ \text{ Only answer possible is } n = 3 &\Leftrightarrow k = 5 \text{ if } k > 2.\\ k = 1,2 \text{ do not satisfy, check individually. } \end{align}