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Let $a$, $b$ and $c$ be the denote $3$ distinct integers and let $P$ denote a polynomial having all integral coefficients. Show that it is impossible that $P(a)=b, P(b)=c, P(c)=a$

My Attempt:

$Q(x)=P(x)-b, R(x)=P(x)-c, S(x)=P(x)-a$

$Q(a)=R(b)=S(c)=0$

$P(x)=(x-a)F(x)=(x-b)G(x)=(x-c)H(x)$

I am stuck here.

amWhy
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Ouestion Asker
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1 Answers1

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If $p(x)$ is a polynomial with integral coefficients then $(a-b)|p(a)-p(b)$. So the problem reduces to prove that if $p(a)=b,p(b)=c$ and $p(c)=a$ then $a=b=c$ where $a,b,c\in I$ .I hope you can take it from here. :)

voidEnergy
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