Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.