a, b and c are integers. If a polynomial exists so p(a)=b, p(b)=c, and p(c)=a

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CancerousProblem
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show that it cannot have integer coefficients
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Zacken
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Proof by contradiction.

Suppose that there does exist a polynomial \displaystyle P(x) with integer coefficients that satisfies the property that \displaystyle P(a) = b, P(b)=c, P(c)=a.

Then \displaystyle a is a root of \displaystyle P(x)-b and using the factor theorem yields \displaystyle P(x)-b = (x-a)Q(x) where \displaystyle Q(x) is a polynomial with integer coefficients.

Specifically, we have that, \displaystyle \frac{c-b}{b-a} = \frac{P(b) -b}{b-a}=Q(b), must be an integer.

Similarly, we can see that both numbers \displaystyle \frac{a-c}{c-b} and \displaystyle \frac{b-a}{a-c} must also be an integer. It is obvious then, that:

\displaystyle \frac{c-b}{b-a} \cdot \frac{a-c}{c-b} \cdot \frac{b-a}{a-c} = 1

This is only possible if \displaystyle |c-b| = |a-b| = |a-c|, which is clearly impossible if \displaystyle a,b \mathrm{and} c are distinct integers. Contradiction.

So there cannot exist a polynomial that cyclically permute \displaystyle a,b \mathrm{and} c.
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