# A problem on powers

I was initially working with a different problem which I solved very easily. Anyway, the problem I'm going to post now, may not be of any particular interest, but I want to have some help towards solving it.

Suppose that $a$, $b$ are two non-negative integers satisfying: $$a^{2n+1}+b^{2n+1}\text{ is a perfect square for all non-negative integers }n$$ (i.e. $a+b$, $a^3+b^3$, $a^5+b^5$, etc. are all perfect squares). Are there any such $a$ and $b$, such that none of $a$ and $b$ is 0?

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And we want to exclude $a=b=2$, and relatives. – André Nicolas Jun 13 '12 at 5:40
Right, Andre Nicolas and Serkan. Exclude the cases a=b=some odd power of 2, which I missed out. So generally speaking, find all possible such a,b. – Somabha Mukherjee Jun 13 '12 at 5:45
For what it's worth, I don't think any examples are known of $a^5+b^5=c^2$ for relatively prime $a,b,c$. The standard conjectures would say there are at most finitely many. This doesn't directly apply to the problem at hand, but it shows what you're up against. – Gerry Myerson Jun 13 '12 at 6:28

It should be possible to find all solutions of the system $$a+b=c^2,\qquad a^3+b^3=d^2$$ Since $a^3+b^3=(a+b)(a^2-ab+b^2)$, these equations imply $$a^2-ab+b^2=e^2$$ for some integer $e$. This equation can be rewritten $$(2a-b)^2+3b^2=(2e)^2$$ which we can think of as $x^2+3b^2=y^2$, and for that equation we can find the general solution by the same techniques used to find the general solution of $r^2+s^2=t^2$. Once you have those solutions, you can go back and see what additional conditions are required to get $a+b$ to be a square.
There are infinitely many positive solutions. For any positive integer $k$ obviously $(a, b) = (2^{2k + 1}, 2^{2k + 1})$ are solutions. There are probably more than that.