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I'm struggling with this question: Suppose $P(x)$ is a polynomial with integer coefficients such that non of the values $P(1),...,P(2010)$ is divisible by $2010$. Prove that $P(n)\neq 0$ for all integers $n$

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up vote 7 down vote accepted

Hint: If $n\equiv m\pmod {M}$, then $P(n)\equiv P(m)\pmod M$.

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Suppose $P(n) = 0$ for some integer $n$. Then write $n = p\cdot 2010 + r$, then $ 0 = P(2010p + r) = 2010q + P(r)$. So $P(r) = -2010q$ which is divisible by $2010$ with $0 \le r < 2010$, contradiction.

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Hint $\ $ A root of $P$ remains a root of $P$ mod $\,2010$ but, by hypothesis, $P$ has no roots mod $\,2010$.

Remark $\ $ More explicitly, mod $\,2010$ every integer $\,n\,$ is congruent to a (unique) integer $\bar n$ in the interval $\,[1,2010].\,$ Indeed, let $\,\bar n\,$ be the remainder mod $2010$ (and replace $\,0\,$ by $\,2010).$ Therefore, if $\,P(n) = 0\,$ then by applying the $\color{#c00}{\rm Polynomial\ Congruence\ Rule}$ (below) we deduce

$$ {\rm mod}\ 2010\!:\,\ \bar n\equiv n\,\Rightarrow\, P(\bar n)\equiv P(n)\equiv 0$$

But this implies that $\,2010\mid P(\bar n)\,$ and $\, 1\le \bar n\le 2010,\ $ contra hypothesis.

Congruence Sum Rule $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#c0f}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#c0f}{A+B - (a+b)} $

Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \ B\equiv b\ \Rightarrow\ \color{blue}{AB\equiv ab}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{blue}{AB - ab} $

Congruence Power Rule $\rm\qquad \color{}{A\equiv a}\ \Rightarrow\ \color{#0a0}{A^n\equiv a^n}\ \ (mod\ m)$

Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#0a0}{A^{n+1}\equiv a^{n+1}},\,$ by the Product Rule, so the result follows by induction on $\,n.$

$\color{#c00}{\rm\bf Polynomial\ Congruence\ Rule}\ $ If $\,f(x)\,$ is polynomial with integer coefficients then $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$

Proof $\ $ By induction on $\, n = $ degree $f.\,$ Clear if $\, n = 0.\,$ Else $\,f(x) = f(0) + x\,g(x)\,$ for $\,g(x)\,$ a polynomial with integer coefficients of degree $< n.\,$ By induction $\,g(A)\equiv g(a)\,$ so $\, \color{#0a0}{A g(A)\equiv a g(a)}\,$ by the Product Rule. Hence $\,f(A) = f(0)+\color{#0a0}{Ag(A)}\equiv f(0)+\color{#0a0}{ag(a)} = f(a)\,$ by the Sum Rule.

Beware $ $ that such rules need not hold true for other operations, e.g. the exponential analog of above $\rm A^B\equiv\, a^b$ is not generally true (unless $\rm B = b,\,$ so it follows by applyimg the Polynomial Rule with $\,f(x) = x^{\rm b}).$

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