Difference between revisions of "1974 USAMO Problems/Problem 1"
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Let <math>a</math>, <math>b</math>, and <math>c</math> denote three distinct integers, and let <math>P</math> denote a polynomial having all integral coefficients. Show that it is impossible that <math>P(a)=b</math>, <math>P(b)=c</math>, and <math>P(c)=a</math>. | Let <math>a</math>, <math>b</math>, and <math>c</math> denote three distinct integers, and let <math>P</math> denote a polynomial having all integral coefficients. Show that it is impossible that <math>P(a)=b</math>, <math>P(b)=c</math>, and <math>P(c)=a</math>. | ||
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+ | == Hint == | ||
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+ | If <math>P</math> is a polynomial with integral coefficients, then <cmath>a - b | P(a) - P(b).</cmath> (Why?) | ||
== Solution == | == Solution == |
Revision as of 15:31, 31 August 2014
Problem
Let , , and denote three distinct integers, and let denote a polynomial having all integral coefficients. Show that it is impossible that , , and .
Hint
If is a polynomial with integral coefficients, then (Why?)
Solution
It suffices to show that if are integers such that , , and , then .
We note that so the quanitities must be equal in absolute value. In fact, two of them, say and , must be equal. Then so , and , so , , and are equal, as desired.
Solution 1b
Let be the value are equal to in absolute value. Assume is nonzero. Then each of is equal to or , so where is one of -3, -1, 1, or 3. In particular, neither nor is zero, contradiction. Hence, , and are equal, as desired.
Solution 2
Consider the polynomial By using the facts that and , we find that Thus, the polynomial has a and b as roots, and we can write for some polynomial . Because and are monic polynomials with integral coefficients, their quotient, , must also have integral coefficients, as can be demonstrated by simulating the long division. Thus, , and hence , must be divisible by . But if and , then we must have, after rearranging terms and substitution, that is divisible by . Equivalently, is divisible by (after canceling the which is clearly divisble by ). 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.