What is the sum of all positive integers $n$ for which $2^n + 65$ is a perfect square?
Perfect square. Factorising problem
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0Aargh, just realized tom hasn't been here in about 10 months. I have to remember to check these things *before* I post.... – 2012-01-22
2 Answers
First show that $n$ cannot be odd. This can be done by looking at $2^n + 65$ modulo 4.
So $n$ must be even, say $n=2m$. If $2^{2m} +65=x^2$, then $65=(x-2^m)(x+2^m)$. Now look at all the factorizations of 65.
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0Every so often I bu$m$p i$n$to a proposed deletion of a quite old question, and therefore of good answers by the founding people of MSE. Shades of Robespierre! – 2014-09-13
HINT
First note that for $2^n + 65$ to be square, $n$ must be even. You can check this since a square $\pmod{10}$ can be $0,1,4,5,6,9$. Hence, $n=2k$. Hence the problem now boils down to finding $k$ such that $4^k + 65$ is a square.
$4^k + 65 = x^2$ i.e. $x^2- (2^k)^2 = 65$ i.e. $(x+2^k)(x-2^k) = 65$.
Now look at the different possible cases of factorizing $65$ over $\mathbb{Z}$.
Analyze the different cases and get the different values of $k$ and add them up. I am unable to think of direct way to evaluate the sum instead of computing the individual $k$'s.