Difference between revisions of "1974 USAMO Problems/Problem 1"

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[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 18:55, 3 July 2013

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 \[0 = \lvert (a-b) + (b-c) + (c-a) \rvert = \lvert 2(a-b) + (c-a) \rvert \ge 2 \lvert a-b \rvert - \lvert c-a \rvert = \lvert a-b \rvert ,\] 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.

See Also

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

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