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Let p(x) be a polynomial over the integers such that it takes 1 value at three different integers. Prove that it has no integral root.

MY ATTEMPT:

I couldn't really think of anything. I first thought that this polynomial cannot be linear or quadratic, as it takes the same values at more than two points(an application of the identity theorem for polynomials) but I dont know where to take that.

2 Answers2

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Let $q(x)=p(x)-1$

$q(a)=q(b)=q(c)=0$ for some integers $a$, $b$, $c$.

$q(x)=(x-a)(x-b)(x-c)\cdot r(x)$, where $r(x)$ is some other polynomial over the integers.

When $p(x)=0$, $(x-a)(x-b)(x-c)\cdot r(x) = -1$

This is impossible since $a,b,c$ are integers, so the closest $(x-a)(x-b)(x-c)$ can be is $2$ or $-2$

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(See my comment for my interpretation of the question.)

Hint: For integer polynomials, $a-b \mid P(a) - P(b)$.

Calvin Lin
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