MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Walter Hayman just asked me the following question. What, if anything, is known about the growth of the function $n(k)$, where $k\geq1$ is an integer, and $n=n(k)\geq2$ is the smallest integer for which there is a non-trivial solution in positive integers to $$x_1^k+x_2^k+\cdots+x_n^k=y^k.$$

Walter observed that an upper bound for $n(k)$ is $G(k)$ [EDIT: this should probably say $G(k)+1$], the "big $G$" function used as standard notation in discussions of Waring's Problem. Explicitly, $G(k)$ is the smallest $m$ such that every sufficiently large integer is the sum of at most $m$ $k$th powers; the trick of course is then to choose a sufficiently large $k$th power [EDIT: and then subtract 1.]. Hence $n(k)$ grows at most as fast as $G(k)$, and a fair bit is known about the growth of $G(k)$ (again see the Wikipedia page); for example $G(k)=O(k\log(k))$. We deduce that $n(k)$ is at most $O(k\log(k))$. The question is whether it is possible to do any better.

I would imagine that lower bounds are hopeless -- it was a lot of work to prove that $n(k)\geq3$ for all $k\geq3$, and I am not sure that mathematics is ready to rule out the possibility that there are arbitrarily large integers $k$ such that there are solutions to $a^k+b^k+c^k=d^k$ in positive integers (i.e. that $n(k)\geq4$ for all sufficiently large $k$). But it did not seem unreasonable to hope that there were better upper bounds than the naive approach going via $G(k)$, so I told him I'd ask here.

share|cite|improve this question
I asked about lower bounds at… – Boris Bukh Mar 16 '12 at 16:45
I don't see why $G(k)$ is an upper bound for $n(k).$ Since every large $k$-th power is already a $k$-th power, I think they are irrelevant to the value of $G(k).$ – Will Jagy Mar 18 '12 at 20:08
$3^3 + 4^3 + 5^3 = 6^3, \; \; \; 1^3 + 6^3 + 8^3 = 9^3, \; \; 3^3 + 10^3 + 18^3 = 19^3, \; \; 7^3 + 14^3 + 17^3 = 20^3$ – Will Jagy Mar 18 '12 at 21:02
Will -- you're right. I guess you can use not $N^k$ but $N^k-1$ in the $G(k)$ result, giving $n(k)\leq G(k)+1$ which is still good enough for what follows. Thanks. I'll fix it. – Kevin Buzzard Mar 19 '12 at 10:26
@Will, isn't D1 more to the point than D7? We have (thanks to Noam Elkies) 3 4th powers that sum to a 4th power, and (thanks to Lander and Parkin) 4 5th powers that sum to a 5th power. Guy says that for $k\ge6$ there is no known sum of $k$ $k$th powers giving a $k$th power. – Gerry Myerson Mar 29 '12 at 23:09

Your Answer


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

Browse other questions tagged or ask your own question.