Why is $f(x) \equiv 0 \;\bmod \; p$ for all $p \in \mathbb{P}$?

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 $1^1+1\equiv0\pmod 3$? – Frank Science Jun 28 '12 at 9:45
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 $\equiv 0$ – ulead86 Jun 28 '12 at 9:48
It's obvious that $0^2+0\equiv0\pmod p$ and $(-1)^2+(-1)\equiv0\pmod p$, otherwise, $x^2+x\not\equiv0\pmod p$. – Frank Science Jun 28 '12 at 9:49
The polynomial $x^2+x=x(x+1)$ has in fact exactly two solutions mod $p$, $x\equiv 0$ and $x\equiv -1$. – Michalis Jun 28 '12 at 9:50
Ah ok, to obvious, but thanks anyway :) – ulead86 Jun 28 '12 at 9:53

$x^2+x=0\pmod p\implies p|x(x+1)$,but since p is a prime $p|x$ or $p|(x+1)$ giving two solutions $x=0\pmod p$ or $x=-1\pmod p$.
You should write congruence modulo p as \pmod{p}, not (\mod p): compare $(\mod p)$ and $\pmod{p}$. – Najib Idrissi Jun 28 '12 at 11:48