Hi verybody,
Is there a proof of:
Sum[(i^s - 1), {i, 1, a}] mod( a ) = Sum[(i - 1)^s, {i, 1, a}] mod(
a )
a, s positive integers (could be equal).
If there is, please tell me.
$\displaystyle
\sum_{i=1}^a(i^s-1) \equiv \Big(\sum_{i=1}^ai^s\Big)-a \equiv \sum_{i=1}^ai^s \mod a$
and
$\displaystyle \sum_{i=1}^ai^s \equiv \sum_{i=1}^a(i-1)^s \mod a$
because both sums are taken over complete residue systems $\displaystyle \mod a$. Works also if $\displaystyle s>a$ b.t.w.