Hard elementary-number-theory question on solve all $n$s that make $2^6+2^{10}+2^n$ a square numbe

Question

I want to know all the nonnegative integer $n$ that makes $2^6+2^{10}+2^n$ be some other integer's square.

I have tried it numerically for a range from $0$ to $1000$, and only $0,9,11,12,15$ returns to be satisfying the condition.

I believe the above is all the solution, but I need a formal proof.

Thanks.

Answer 1

This is not a direct answer to your question, but it may help analyzing the problem:

$\exists n>15$ such that $2^6+2^{10}+2^n$ is a perfect square $\iff$

$\exists n>15$ such that $2^6(2^0+2^4+2^{n-6})$ is a perfect square $\iff$

$\exists n>15$ such that $64(17+2^{n-6})$ is a perfect square $\iff$

$\exists n>15$ such that $17+2^{n-6}$ is a perfect square $\iff$

$\exists m>9$ such that $17+2^m$ is a perfect square

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So an equivalent but simplified question would be:

Is $17+2^m$ a perfect square only for a positive integer $m=3,5,6,9$?

Answer 2

$2^n$ is either a square, $p^2$, or twice a square, $2q^2$, depending on whether $n$ is even or odd.

17 can only be the difference between two squares in a small number of ways.

So you are left solving the Pell-type equation $2q^2+17=r^2$