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

Revision as of 17:55, 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.

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 {{USAMO box|year=1974|beforetext=|before=First Question|num-a=2}}
 * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=579731#579731 Discussion on AoPS/MathLinks]
+{{MAA Notice}}
 [[Category:Olympiad Algebra Problems]]

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

Revision as of 17:55, 3 July 2013

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