Suppose $f(x)$ is a polynomial with integer coefficients. Show that if $f(a)=p$ for some integer $a$ and prime number $p$, then $f(x)$ has at most three integer roots; that is, there are at most three distinct integers $\alpha,\beta,\text{ and }\gamma$ such that $f(\alpha)=f(\beta)=f(\gamma) = 0$.
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So $f(a)=p$ and $f(a)=0$? – Angina Seng Feb 01 '18 at 05:42
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you used $a$ twice – Sean Nemetz Feb 01 '18 at 05:43
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Please use MathJax to format and show some efforts. – Ѕᴀᴀᴅ Feb 01 '18 at 05:46
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1Think about the factorisation of $f$. – Angina Seng Feb 01 '18 at 05:46
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1If $p=3$ there is an answer here. If $p$ is any other prime you can easily modify this answer. – David Feb 01 '18 at 06:01
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David points out the answer in a comment. Let me adapt that answer here:
Suppose $f(x)$ has four roots $\alpha,\beta,\gamma$ and $\delta$. Then $f(x)=g(x)(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$. Applying to $a$, we find $$ g(a)(a-\alpha)(a-\beta)(a-\gamma)(a-\delta)=p. $$ But $(a-\alpha),(a-\beta),(a-\gamma),(a-\delta)$ are four distinct integers, and this is impossible. This is because if $p$ is prime, the maximal amount of distinct integers we can factor it into is $3$, given by multiplying its divisors: $$ p=(-p)\cdot(-1)\cdot 1$$
Therefore $f$ has at most $3$ roots.
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