Take the 2-minute tour ×  Hi everyone, I got a problem when
proving lemmas for some combinatorial problems, and it is a
question about integers.   $\sum_{k=1}^m a_k^t = \sum_{k=1}^n
b_k^t$  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 soluti