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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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    What about $1^1+1\equiv0\pmod 3$?2012-06-28
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    If you set $x=2$ you'll get 0. I tried to say it don't must be true for all x's, but there must be at least one x mod p which gives $\equiv 0$2012-06-28
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    It's obvious that $0^2+0\equiv0\pmod p$ and $(-1)^2+(-1)\equiv0\pmod p$, otherwise, $x^2+x\not\equiv0\pmod p$.2012-06-28
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    The polynomial $x^2+x=x(x+1)$ has in fact exactly two solutions mod $p$, $x\equiv 0$ and $x\equiv -1$.2012-06-28
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    Ah ok, to obvious, but thanks anyway :)2012-06-28
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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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    You should write congruence modulo p as \pmod{p}, not (\mod p): compare $(\mod p)$ and $\pmod{p}$.2012-06-28
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    @N.I:Thanks for the suggestion.2012-06-28