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Can we find all positive integers $a,b$ such that $a^{n}+b^{n}$ is an $(n+1)^{th}$ power? I think this question equivalent to solving the statement $$a^{n} + b^{n} = c^{n+1}$$ for $a,b,c \in \mathbb{N}$. But i don't know as to how i can solve this.

I attempted by subsituting $n$ as $n+1$ so that we get $$a^{n+1} + b^{n+1} = c^{n+2} = c \times c^{n+1}=(a^{n}+b^{n}) \cdot c$$

but it seems that this doesn't help!

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I don't understand the substitution. You seem to be trying to construct solutions inductively but what you're actually doing is forcing a, b, c to satisfy both the equation for n and for n+1. –  Qiaochu Yuan Oct 15 '10 at 23:34
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4 Answers

up vote 8 down vote accepted

When $n = 2$ it's straightforward to give a more-or-less complete solution: $c^3$ is a sum of squares if and only if $c$ is, so write $c = d^2 + e^2$ and the rest is factorization over the Gaussian integers.

For all $n$ there is the family of solutions $a = x(x^n + y^n), b = y(x^n + y^n), c = x^n + y^n$ for integers $x, y$. But I think finding all solutions is likely to be incredibly hard.

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In particular, finding all relatively prime solutions seems extremely hard to me. Even for particular small values of n this is probably not known. –  Qiaochu Yuan Oct 15 '10 at 23:33
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For $n \ge 3$, the Fermat-Catalan conjecture (http://en.wikipedia.org/wiki/Fermat–Catalan_conjecture) implies that there are only finitely many solutions to your equation. (But the Fermat-Catalan conjecture is just a conjecture!)

Edit: as pointed out by yrudoy, this doesn't actually follow. I still think there should be only finitely many solutions, though, based on some probabilistic considerations. See for example Noam Elkies, The ABCs of Number Theory, in vol. 1, issue 1 of the Harvard College Mathematics Review, p. 61. (This gives a standard heuristic argument for Fermat's last theorem only having finitely many solutions for each $n \ge 4$, which adapts to this case.

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Actually, this conjecture requires b,c, and d to be coprime, which is not necessarily the case. –  yrudoy Oct 16 '10 at 4:50
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The hard case is relatively prime solutions. For $n > 2$ it is conjectured that there are no solutions. This follows from the more precise conjecture that there are no relatively prime solutions to $x^a + y^b = z^c$ in the hyperbolic case $1/a + 1/b + 1/c < 1$ other than the ten or so examples found by Beukers and Zagier in a computer search (there are large solutions with $(a,b,c)=(2,3,7)$ and a few others) shortly after Wiles announced the proof of Fermat's Last Theorem. It was proven by Darmon and Granville that the number of relatively prime solutions is finite for any given $n > 2$, and this required the Mordell conjecture (Faltings' theorem) on finiteness of rational points on curves of genus $g \geq 2$. There is no known elementary approach to these questions, and resolution of the problem for particular values of $(a,b,c)$ uses the full machinery of modern algebraic geometry and number theory.

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A few to start you off: for n=1 there are many solutions. a=b=c=2 works for all n. There is http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares. You can apply that to c^3 to see if all factors of c which are 3 mod 4 come with even exponents you can solve it. I suspect that is it as there just aren't that many powers to choose from, but don't have a proof.

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