Find all positive integers $n$ such that $2^8+2^{12}+2^n$ is a perfect square

Question

Find all positive integers $n$ such that $2^8+2^{12}+2^n$ is a perfect square.

For $n=2$ and $n=11$, $2^8+2^{12}+2^n$ is a perfect square.

How to find a closed form?

Answer 1

If $n\in\{1,2,\ldots,7\}$, then $n=2$ is the only solution. Let $n\ge 8$. Then $2^8\left(1+2^4+2^{n-8}\right)$ is a square iff $17+2^{n-8}$ is a square. Let $n-8=m$. You're solving $17+2^m=k^2$ in non-negative integers.

All the solutions are $m=3,5,6,9$. Now, the solutions for $m$ are relatively large, and assuming $m\ge 10$ and using mod $2^{k}$, $k\ge 10$ won't help, because $17$ will always be a quadratic residue.

This is very similar to Ramanujan–Nagell equation, namely $2^n-7=x^2$, which has the solutions $n=3,4,5,7,15$.

This equation has already been asked about on M.SE (see here and here), and the only solution someone could come up with uses Mordell's equations, so I'll use them here. We have three cases:

• $m=3t,\, t\ge 0$. Then $17+\left(2^t\right)^3=k^2$. But $17+a^3=b^2$ has $16$ solutions (see http://oeis.org/A081119 and http://oeis.org/A081119/b081119.txt), which can be found with a program or this table:

  $$(a,b)=(-2,\pm 3), (-1,\pm 4), (2,\pm 5), (4,\pm 9),$$

  $$ (8,\pm 23), (43,\pm 282), (52,\pm 375), (5234,\pm 378661)$$ This gives $m=3,6,9$.

• $m=3t+1,\, t\ge 0$. Then $2^m\equiv 2\pmod{7}$, so $k^2\equiv 5\pmod{7}$, impossible, because $5$ is not a quadratic residue mod $7$.

• $m=3t+2,\, t\ge 0$. Then $272+\left(2^{t+2}\right)^3=(4k)^2$. But $272+a^3=b^2$ has $2$ solutions (http://oeis.org/A081119/b081119.txt), which are $(a,b)=(8,\pm 28)$. This gives $m=5$.