Prove that there is no polynomial with integer coefficients such that $p(a)=b,\,p(b)=c,\,p(c)=a$ for distinct integers $a,b,c$

Question

Our teacher gave us this question but I am very stuck. I drew graphs to see why it cant be true but I didnt find anything. I see that if $p$ existed then: $$p(\cdots p(a))=a,\;p(\cdots p(b))=b,\;p(\cdots p(c))=c$$

(With $3n$ $p$'s). But I don't know where the contradiction is... Maybe we can work on the divisibility of the polynomials coefficients or something?

The integer coefficients condition is important. Recall that $x - y | p(x) - p(y)$. — Qiaochu Yuan, Mar 15 '13 at 21:00
If we could integers mod 3, then x+1 would work actually as p(0)=1,p(1)=2,p(2)=0. Thus one has to be careful about what domain is one using here. — JB King, Mar 16 '13 at 05:28

Math Gems · Accepted Answer · 2013-03-15T22:13:57.527

1

Hint $\ $ Writing $\rm\,px\,$ for $\rm\,p(x)\,$ and applying the Factor Theorem we have

$$\rm\,\ a\!-\!b\mid pa\!-\!pb = b\!-\!c\mid pb\!-\!pc=c\!-\!a\mid pc\!-\!pa = a\!-\!b\ $$

therefore we deduce $\rm\,\ a\!-\!b\mid b\!-\!c\mid c\!-\!a\mid a\!-\!b\ $ hence $\rm\ \ldots$

edited Mar 15 '13 at 22:13

answered Mar 15 '13 at 21:37

Math Gems

19,574

This would be much clearer if you wrote p(a) for pa, etc. – Chris Godsil Mar 15 '13 at 21:58
@Chris But all the parens would make the structure less clear. – Math Gems Mar 15 '13 at 22:14
I can see your point, but it did confuse me for a bit. – Chris Godsil Mar 16 '13 at 01:31
@Chris That's why I prepended a notational remark. Thanks for the feedback. – Math Gems Mar 16 '13 at 01:45

Prove that there is no polynomial with integer coefficients such that $p(a)=b,\,p(b)=c,\,p(c)=a$ for distinct integers $a,b,c$

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