Equality of the sum of powers 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 $n = m$?

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.)
