For how many of value(s) of n, $n \in \mathbb{N}$, $2^8 + 2^{11} + 2^{2n}$ is a perfect square? [duplicate]

Possible Duplicate:
Two problems on number theory

For how many of value(s) of n, $n \in \mathbb{N}$, $2^8 + 2^{11} + 2^{2n}$ is a perfect square?

How to do it?

-

marked as duplicate by Gerry Myerson, t.b., Asaf Karagila, Zev ChonolesJan 19 '12 at 17:17

One approach is just to try. Can you see why you don't have to test $n > 11$? –  Ross Millikan Nov 2 '11 at 22:18
We break the problem up into two parts. First we deal with the "open-ended" part, where $n\ge 6$. In this case, our expression is equal to $$2^8(1+2^3+2^{2n-8}).$$ Since $2^8$ is a perfect square, we need $2^{2n-8} +9$ to be a perfect square. We have the solution $2n-8=4$. There are no other solutions with $n\ge 6$. For if $x^2$ is a perfect square, the next perfect square is $x^2+2x+1$. So if $2^n$ is bigger than $4$, then the next perfect square after $2^{2n}$ is $(2^{2n}+1)^2$, which is bigger than $2^{2n}+9$. We conclude that there is exactly one solution with $n \ge 6$.
Next we look at the cases $n=1$, $2$, $3$, $4$, and $5$ (and perhaps $0$ if we allow that). There is no sense in thinking too much about the matter: we can just use the calculator.
Comment: The simple observation that if $x$ is an integer, then there is no perfect square strictly between $x^2$ and $(x+1)^2$ is surprisingly useful in Diophantine problems.