1974 USAMO Problems/Problem 1

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. $\blacksquare$

Solution 2

Consider the polynomial $Q(x) = (b-a)P(x) - (c-b)x - b^2 + ac.$ By using the facts that $P(a) = b$ and $P(b) = c$ , we find that $Q(a) = Q(b) = 0.$ Thus, the polynomial $Q(x)$ has a and b as roots, and we can write $Q(x) = (x-a)(x-b)R(x)$ for some polynomial $R(x)$ . Because $Q(x)$ and $(x-a)(x-b)$ are monic polynomials with integral coefficients, their quotient, $R(x)$ , must also have integral coefficients, as can be demonstrated by simulating the long division. Thus, $Q(c)$ , and hence $Q(c) - (x-a)(x-b)$ , must be divisible by $(c-a)(c-b)$ . But if $x=c-a$ and $y=c-b$ , then we must have, after rearranging terms and substitution, that $(x-y)^2$ is divisible by $xy$ . Equivalently, $S = x^2 + y^2$ is divisible by $xy$ (after canceling the $2xy$ which is clearly divisble by $xy$ ). Because S must be divisible by both x and y, we quickly deduce that x is divisible by y and y is divisible by x. Thus, x and y are equal in absolute value. A similar statement holds for a cyclic pemutation of the variables, and so x, y, and z are all equal in absolute value. The conclusion follows as in the first solution.