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Let $p(x)$ be a polynomial with integral coefficients. Let $a$, $b$, $c$ be three distinct integers such that $p(a) = p(b) = p(c) = -1$. Find the number of integral roots of $p(x)$.

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Any condition about $\deg p$? –  Hagen von Eitzen Dec 24 '13 at 10:04
    
What can you say about the structure of a polynomial with $p(a) = p(b) = p(c) = -1$? –  Daniel Fischer Dec 24 '13 at 10:05
    
Is it,find the number of integral roots or the minimum number of integral roots? –  GTX OC Dec 24 '13 at 10:36
    
No there is nothing about the polynomial... –  posthumus Dec 24 '13 at 10:45
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3 Answers

Suppose that $p(d)=0$ for some integer $d$. Then $(x-d) \mid p(x)$. Plugging in, we discover a very surprising fact about the integers $a-d$, $b-d$, and $c-d$...

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I did not get it, sorry –  posthumus Dec 24 '13 at 10:40
    
@posthumus: Since $(x-d)\mid p(x)$ it follows that $(a-d)\mid p(a)=-1$. Therefore $(a-d)\mid -1$ and similarly $(b-d)\mid -1, (c-d)\mid -1$. Also $a-d, b-d, c-d$ are $3$ different integers. Now how many factors does $-1$ have? –  P.. Dec 24 '13 at 11:03
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Do you know factor theorem?

$p(a)=-1$ implies that the remainder when you divide $p(x)$ by $x-a$ is $-1$.

Using this theorem gives you a representation of $p(x)$.

Then, you'll know a 'surprising' fact (as User-33433 says) when you set an integer $x=d$ in $p(x)$.

By factor theorem, we get $$p(x)=(x-a)(x-b)(x-c)q(x)-1$$ where $q(x)$ is a polynomial with integral coefficients.

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And now the interesting fact is that since there are at least three numbers such that $p(x)=-1$, you know for sure that one of $|d-a|$, $|d-b|$, $|d-c|$ is greater than $1$. Thus $p(d)=kq(d)-1=0$ would imply $q(d)=1/k$, witk $|k|>1$, but $d$ is assumed to be an integer, and $q$ has integer coefficients. –  Jean-Claude Arbaut Dec 24 '13 at 11:05
    
That's right. Hence...nice question. –  mathlove Dec 24 '13 at 11:13
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It deserves explicit emphasis that there is a frequently applicable idea at work here.

Key Idea $\ $ The possible factorizations of a polynomial $\in\Bbb Z[x]$ are constrained by the factorizations of the integer values that the polynomial takes. For a simple example, if some integer value has few factorizations (e.g. a unit $\,\pm1 $ or prime $p$) then the polynomial must also have few factors, asssuming that that the factors are distinct at the evaluation point. More precisely

If $\, f(x) = f_1(x)\cdots f_k(x)\,$ and $\,f_i\in\Bbb Z[x]\,$ satisfy $\color{#0a0}{f_i(n) \ne f_j(n)}\,$ for $\,i\ne j,$ all $\,n\in \Bbb Z$

$\quad f(n) =\pm1\,\Rightarrow\, k\le 2\ $ else $1$ would have $\rm\,3\,\ \color{#0a0}{distinct}$ factors $\,f_1(n),f_2(n),f_3(n)$

$\quad f(n) = \pm p\,\Rightarrow\, k\le \color{#c00}4\ $ since a prime $p$ has at most $\,\color{#c00}4\,$ distinct factors $\,\pm1,\pm p$

Yours is a special case of the first (unit) case, where the $f_i$ are linear.

Remark $\ $ One can push the key idea to the hilt to obtain a simple algorithm for polynomial factorization using factorization of its integer values and Lagrange interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker. See this answer for references.

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