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, 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.