Let $a,b,c$ be 3 distinct integers, $P(x)$ is a polynomial with integer coefficient, prove that we cannot have $P(a)=b,P(b)=c,$ and $P(c)=a$(USAMO1974)
Prove. we assume that there exsits a $P(x)$ satisfied the conditions. Notice that $a-b|P(a)-P(b)=b-c$ by the same method, we have
$a-b|b-c,b-c|c-a,c-a|a-b$ so$|a-b|=|b-c|=|c-a|$, we only need to consider the situation without abs(because other case are obviously contradiction) $a-b\le b-c,b-c\le c-a, c-a\le a-b$, but $a-b+b-c+c-a=0$, therefore we have $a-b=b-c=c-a$
let $b=c+d,a=c+2d$,we have$c-a=-2d=d$ hence $d=0\iff a=b=c $ which is a contradiction.
No comments:
Post a Comment