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I try to find a reason/proof for the following statement: Let be $f(x)=x^2+x$ an integer polynomial. Why is $x^2+x \equiv 0 \pmod p$ for all $p \in \mathbb{P}$?

I made a list for the first primes and obviously it's true, but I can't find a proof for it.

Any help would be great.

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    That's not true. Perhaps you intended $\:x^p - x \equiv 0\pmod p,\:$ i.e. Fermat's Little Theorem.2012-06-28

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$x^2+x=0\pmod p\implies p|x(x+1)$,but since p is a prime $p|x$ or $p|(x+1)$ giving two solutions $x=0\pmod p$ or $x=-1\pmod p$.

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    @N.I:Thanks for the suggestion.2012-06-28