1974 USAMO Problems/Problem 1
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, yielding a final contradiction of the existence of these variables.
In fact, this approach generalizes readily. Suppose that is an odd integer, and that there exist distinct integers such that and , for . Then we have so the differences are all equal in absolute value and thus equal to or for some integer . Adding the differences (in the order they are written) gives for some odd integer , so is nonzero and hence . Thus, all the are equal, a contradiction of their distinctness, so the sequence cannot exist.
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.