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

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===Solution 1b===
 
===Solution 1b===
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Let <math>k</math> be the value <math>(a-b), (b-c), (c-a)</math> are equal to in absolute value. Assume <math>k</math> is nonzero. Then each of <math>(a-b), (b-c), (c-a)</math> is equal to <math>k</math> or <math>-k</math>, so <cmath>0 = (a-b) + (b-c) + (c-a) = mk,</cmath> where <math>m</math> is one of -3, -1, 1, or 3. In particular, neither <math>m</math> nor <math>k</math> is zero, contradiction. Hence, <math>k = 0</math>, and <math>a, b, c</math> are equal, as desired. <math>\blacksquare</math>
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Let <math>k</math> be the value <math>(a-b), (b-c), (c-a)</math> are equal to in absolute value. Assume <math>k</math> is nonzero. Then each of <math>(a-b), (b-c), (c-a)</math> is equal to <math>k</math> or <math>-k</math>, so <cmath>0 = (a-b) + (b-c) + (c-a) = mk,</cmath> where <math>m</math> is one of -3, -1, 1, or 3. In particular, neither <math>m</math> nor <math>k</math> is zero, contradiction. Hence, <math>k = 0</math>, and <math>a, b, c</math> are equal, yielding a final contradiction of the existence of these variables. <math>\blacksquare</math>
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In fact, this approach generalizes readily. Suppose that <math>n</math> is an odd integer, and that there exist <math>n</math> distinct integers <math>a_1, a_2, ..., a_n</math> such that <math>P(a_i) = a_{i+1}</math> and <math>P(a_n) = 1</math>, for <math>1 \le i \le n-1</math>. Then we have <cmath>a_1 - a_2 | a_2 - a_3 | ... | a_n - a_1 | a_1 - a_2,</cmath> so the differences are all equal in absolute value and thus equal to <math>k</math> or <math>-k</math> for some integer <math>k</math>. Adding the differences (in the order they are written) gives <math>0 = mk</math> for some odd integer <math>m</math>, so <math>m</math> is nonzero and hence <math>k = 0</math>. Thus, all the <math>a_i</math> are equal, a contradiction of their distinctness, so the sequence <math>a_i</math> cannot exist.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 15:37, 31 August 2014

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 $P(c)=a$.


Hint

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

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 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.


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

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