2010 AIME II Problems/Problem 10
Problem
Find the number of second-degree polynomials with integer coefficients and integer zeros for which
.
Solution
Solution 1
Let . Then
. First consider the case where
and
(and thus
) are positive. There are
ways to split up the prime factors between
,
, and
. However,
and
are indistinguishable. In one case,
, we have
. The other
cases are double counting, so there are
.
We must now consider the various cases of signs. For the cases where
, there are a total of four possibilities, For the case
, there are only three possibilities,
as
is not distinguishable from the second of those three.
Thus the grand total is .
Solution 2
We use Burnside's Lemma. The set being acted upon is the set of integer triples such that
. Because
and
are indistinguishable, the permutation group consists of the identity and the permutation that switches
and
. In cycle notation, the group consists of
and
. There are
fixed points of the first permutation and
fixed points of the second permutation. By Burnside's Lemma there are
distinguishable triples
.
Note: The permutation group is isomorphic to .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 11 | |
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