1974 USAMO Problems/Problem 1

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Problem

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

Solution

It suffices to show that if $a,b,c$ are integers such that $P(a) = b$ , $P(b)=c$ , and $P(c)= a$ , then $a=b=c$ .

We note that $a-b \mid P(a) - P(b) = b-c \mid P(b)-P(c) = c-a \mid P(c) - P(a) = a-b ,$ so the quanitities $(a-b), (b-c), (c-a)$ must be equal in absolute value. In fact, two of them, say $(a-b)$ and $(b-c)$ , must be equal. Then $0 = \lvert (a-b) + (b-c) + (c-a) \rvert = \lvert 2(a-b) + (c-a) \rvert \ge 2 \lvert a-b \rvert - \lvert c-a \rvert = \lvert a...$ so $a=b= P(a)$ , and $c= P(b) = P(a) = b$ , so $a$ , $b$ , and $c$ are equal, as desired. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1974 USAMO (Problems)
First Problem	1 • 2 • 3 • 4 • 5	Followed by Problem 2
All USAMO Problems and Solutions

Discussion on AoPS/MathLinks

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1974 USAMO Problems/Problem 1

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