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From Faulhaber's formula we can see that $S(x,y)*(y+1)*lcm _{2i\leq y} den (B_{2i})$ is divisible by $x$, so for $x>z*(y+1)*lcm _{2i\leq y} den (B_{2i})$ our expression $S(x,y)$ gives remainder more than $z$ modulo $x$ provided $z\not =0$.