Equality of the sum of powers

Question

Hi everyone, I got a problem when proving lemmas for some combinatorial problems, and it is a question about integers.

Let

$\sum_{k=1}^m a_k^t = \sum_{k=1}^n b_k^t$

be an equation, where $m, n, t, a_i, b_i$ are positive integers, and $a_i \neq a_j$ for all $i, j$, $b_i \neq b_j$ for all $i, j$, $a_i \neq b_j$ for all $i, j$.

Does the equality have no solutions?

For $n \neq m$, it is easy to find solutions for $t=2$ by Pythagorean theorem, and even for $n = m$, we have solutions like

$1^2 + 4^2 + 6^2 + 7^2 = 2^2 + 3^2 + 5^2 + 8^2$.

For $t > 2$, similar equalities hold:

$1^2 + 4^2 + 6^2 + 7^2 + 10^2 + 11^2 + 13^2 + 16^2 = 2^2 + 3^2 + 5^2 + 8^2 + 9^2 + 12^2 + 14^2 + 15^2$ and $1^3 + 4^3 + 6^3 + 7^3 + 10^3 + 11^3 + 13^3 + 16^3 = 2^3 + 3^3 + 5^3 + 8^3 + 9^3 + 12^3 + 14^3 + 15^3$,

and we can extend this trick to all $t > 2$.

The question is, if we introduce one more restriction, that is, $|a_i - a_j| \geq 2$ and $|b_i - b_j| \geq 2$ for all $i, j$, is it still possible to find solutions for the equation?

For $t = 2$ we can combine two Pythagorean triples, say,

$5^2 + 12^2 + 25^2 = 7^2 + 13^2 + 24^2$,

but how about the cases for $t > 2$ and $n = m$?

Answer 1

An even harder problem than $t>2$ and $n=m$ is the Prouhet–Tarry–Escott problem. Now I leave it to you and google to find lots of examples ;-)

http://en.wikipedia.org/wiki/Prouhet-Tarry-Escott_problem

Answer 2

One set of solutions for t = 3 is the class of numbers known as Taxicab Numbers, named after the number of a taxicab G. H. Hardy took, 1729, that Ramanujan mentioned was equal to 1³ + 12³ = 9³ + 10³. This particular example fails, as |10 - 9| = 1 < 2, but there are other Taxicab numbers, such as:
167³ + 436³ = 228³ + 423³ = 255³ + 414³.

This might be a helpful site for your question.

-Gabriel Benamy

Answer 3

For any $t$, if $m$ is sufficiently large relative to $t$, and $n$ is any positive integer (possibly equal to $m$), then the circle method proves that there exists an infinite sequence of increasingly large solutions such that the ratios between the $a_1,\ldots,b_n$ approach any real positive ratios you want (assuming that a real solution with those ratios exists for that $m,n,t$). This answers your question with much stronger inequalities than the ones you imposed. If you want, you can simultaneously specify the residues of the $a_i$ and $b_j$ modulo some fixed number $N$, provided that those residues are compatible with the equation. In fact, again when $m \gg t$, you can even fix the $b_j$ in advance: the solution to Waring's problem guarantees the existence of $a_1,\ldots,a_m$, and a slight strengthening lets you impose your inequalities too, if the $b_j$ are large enough. (Proof: the circle method again.)