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 $n = m$?



Equality of the sum of powers
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>


