Difference between revisions of "1974 USAMO Problems/Problem 1"
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+ | ==Solution 2== | ||
+ | Consider the polynomial <math>Q(x) = (b-a)P(x) - (c-b)x - b^2 + ac.</math> By using the facts that <math>P(a) = b</math> and <math>P(b) = c</math>, we find that <math>Q(a) = Q(b) = 0.</math> Thus, the polynomial <math>Q(x)</math> has a and b as roots, and we can write <math>Q(x) = (x-a)(x-b)R(x)</math> for some polynomial <math>R(x)</math>. Because <math>Q(x)</math> and <math>(x-a)(x-b)</math> are monic polynomials with integral coefficients, their quotient, <math>R(x)</math>, must also have integral coefficients, as can be demonstrated by simulating the long division. Thus, <math>Q(c)</math>, and hence <math>Q(c) - (x-a)(x-b)</math>, must be divisible by <math>(c-a)(c-b)</math>. But if <math>x=c-a</math> and <math>y=c-b</math>, then we must have, after rearranging terms and substitution, that <math>(x-y)^2</math> is divisible by <math>xy</math>. Equivalently, <math>S = x^2 + y^2</math> is divisible by <math>xy</math> (after canceling the <math>2xy</math> which is clearly divisble by <math>xy</math>). 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. | ||
Revision as of 00:12, 4 August 2014
Contents
Problem
Let , , and denote three distinct integers, and let denote a polynomial having all integral coefficients. Show that it is impossible that , , and .
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 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.