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Let $x,y,z$ are integer and $x,y>0$

Define $S(x,y)=1^y+2^y+3^y+...+x^y$

Can it be shown that

If given $z\ne0$ then $S(x,y)\equiv z\pmod{x}$ have finitely many solution of $x$ with respect to $y$.

Example

Let $z=\pm1$ and $y\equiv 1\pmod2$ then $x=1,2$

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  • $\begingroup$ "...have finitely many solution of $x$ with respect to $y$." What does this mean? $\endgroup$
    – Wojowu
    Commented Apr 11, 2020 at 17:24
  • $\begingroup$ @Wojowu $y$ and $z$ given then there are finitely many $x$ satisfy $S(x,y)\equiv z\pmod{x}$. $\endgroup$
    – Pruthviraj
    Commented Apr 11, 2020 at 17:45
  • $\begingroup$ This is Faulhaber’s Formula which include Bernouli numbers such that are congruent ot Z and fermat little theorem w'd be work here as well $\endgroup$
    – zeraoulia rafik
    Commented Apr 12, 2020 at 2:10

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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 $z\not =0$.

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