1974 USAMO Problems/Problem 1


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 $P(c)=a$.


If $P$ is a polynomial with integral coefficients, then \[a - b | P(a) - P(b).\] (Why?)


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 1b

Let $k$ be the value $(a-b), (b-c), (c-a)$ are equal to in absolute value. Assume $k$ is nonzero. Then each of $(a-b), (b-c), (c-a)$ is equal to $k$ or $-k$, so \[0 = (a-b) + (b-c) + (c-a) = mk,\] where $m$ is one of -3, -1, 1, or 3. In particular, neither $m$ nor $k$ is zero, contradiction. Hence, $k = 0$, and $a, b, c$ are equal, yielding a final contradiction of the existence of these variables. $\blacksquare$

In fact, this approach generalizes readily. Suppose that $n$ is an odd integer, and that there exist $n$ distinct integers $a_1, a_2, ..., a_n$ such that $P(a_i) = a_{i+1}$ and $P(a_n) = 1$, for $1 \le i \le n-1$. Then we have \[a_1 - a_2 | a_2 - a_3 | ... | a_n - a_1 | a_1 - a_2,\] so the differences are all equal in absolute value and thus equal to $k$ or $-k$ for some integer $k$. Adding the differences (in the order they are written) gives $0 = mk$ for some odd integer $m$, so $m$ is nonzero and hence $k = 0$. Thus, all the $a_i$ are equal, a contradiction of their distinctness, so the sequence $a_i$ cannot exist.

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.

Solution 3

Assume for the sake of contradiction that is possible for unique integers $a,b,c$. Let $P(x)=d_1x^n+d_2x^{n-1}+\cdots+d_n.$ Note that \[b=P(a)=d_1a^n+d_2a^{n-1}+\cdots+d_n\] \[c=P(b)=d_1b^n+d_2b^{n-1}+\cdots+d_n\] \[a=P(c)=d_1c^n+d_2c^{n-1}+\cdots+d_n\] Subtracting the second from the first, third from second, and first from third gives \[b-c=P(a)-P(b)=d_1(a^n-b^n)+d_2(a^{n-1}-b^{n-1})+\cdots+ d_{n-1}(a-b)\] \[c-a=P(b)-P(c)=d_1(b^n-c^n)+d_2(b^{n-1}-c^{n-1})+\cdots +d_{n-1}(b-c)\] \[a-b=P(c)-P(a)=d_1(c^n-a^n)+d_2(c^{n-1}-a^{n-1})+\cdots +d_{n-1}(c-a)\] By the RHS, note that $a-b\mid P(a)-P(b)$ so $\lvert{a-b}\rvert\leq\lvert{b-c}\rvert.$ Similarly, $\lvert{b-c}\rvert\leq\lvert{c-a}\rvert$ and $\lvert{c-a}\rvert\leq\lvert{a-b}\rvert.$ Hence, $\lvert{a-b}\rvert\leq\lvert{b-c}\rvert\leq\lvert{c-a}\rvert\leq\lvert{a-b}\rvert$ so $\lvert{a-b}\rvert=\lvert{b-c}\rvert=\lvert{c-a}\rvert.$ Assume WLOG that $a>b>c$ so $a-b=a-c$ and $b-c=c-a.$ From the first equation, we get $b=c$ and substituting this in the second gives $c=a.$ Hence, $a=b=c$, contradicting the uniqueness of $a,b,c.$ $\blacksquare$

Solution 4

Assume that $a, b$ and $c$ are distinct integers which satisfy the given conditions. Then we have $P(c) = P(P(b)) = P(P(P(a))) = a$. Similarly $P(P(P(b))) = b$ and $P(P(P(c))) = c$. Since the polynomial has integer coefficients, this implies that the polynomial $P$ is $P(x) = x$. But this is a contradiction since then $P(a) = b$ can only be true if $a = b$ which contradicts the fact that $a, b$ and $c$ are distinct integers.

Solution 5

We get that $a-b|b-c, b-c|c-a, c-a|a-b$. Thus, we can say that $(b-c) = (a-b)k, (c-a) = (b-c)j, (a-b) = (c-a)m$, for non zero integers, not necesarily positive, k, j, and m. Multiplying these equations and canceling, we get that $kjm$=1. This means that either k=1, m=1, j=1, or two of the variables are equal to 1 and the other is equal to -1. If all of the variables are equal to zero, plugging in gives us that $a=b=c$, contradictory to the conditions given in the problem statement. If two of the variables are equal to -1, and the other is equal to 1, then WLOG let $k=m=-1, j=1$. But this implies that the distinct integer condition is not satisfied, and hence, this case is invalid. Thus, it is impossible. $\blacksquare$ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS