Find the solutions in positive integers $m^2 + 615 =2^n$

Question

Find all positive integer solutions $m$ and $n$ of the equation $$ m^2 + 615 =2^n. $$

Answer 1

Note that $3 \mid 615$ and $3$ is never a factor of $2^n$, so $3 \nmid m$.

Therefore $m^2 \equiv 1 \bmod 3$ and thus also $2^n\equiv 1 \bmod 3$.

Odd powers of $2$ are $\equiv 2 \bmod 3$ and even powers of $2$ are $\equiv 1 \bmod 3$, so we know that $n$ is even. So say $n=2k$.

Now we have $m^2 +615 =2^{2k}$ so $615= 2^{2k}-m^2 = (2^k)^2-m^2 = (2^k+m)(2^k-m)$

The factor pairs of $615$ are $\{(1,615),(3,205),(5,123),(15,41)\}$ so the possible values of $2^k$ from those are $\{308,104,64,28\}$. Only $64$ is a power of $2$ among these, so this gives $k=6$ and

$\hspace{2in}\boxed{ n=12 \\ m=123-64=59}$

Check: $59^2+615 = 3481+615= 4096=2^{12} \quad\bigstar$