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.
You may ask: How can one of be positive and the other negative? will be negative as a result. That way, it's still that gets multiplied.
Thus the grand total is .
Note: The only reason why we can be confident that is the only case where the polynomials are being overcounted is because of this: We have the four configurations listed below:
And notice, we start by counting all the positive solutions. So and must be strictly positive, no or negatives allowed. The negative transformations will count those numbers.
So with these we can conclude that only the first and second together have a chance of being equal, and the third and fourth together. If we consider the first and second, the term would have coefficients that are always different, and because of the negative and . Since the is never equal, these can never create equal coefficients. We don't need to worry about this as and are positive and so that won't have any chance.
However with the and , we have the coefficients of the term as and . In other words, they are equal if or . Well if , then we have and in the case we have and if we transform using , then we have . So this is the only way that we could possibly overcount the equal cases, and so we need to make sure we don't count and twice as they will create equal sums. This is why we subtract from .
Each different transformation will give us different coordinates it is just that some of them create equal coefficients for the -term, and we see that they are equal only in this case by our exploration, so we subtract to account and get .
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 (after distributing the primes among , , and then considering their signs. We have 4 ways since we can keep them all positive, first 2 negative, first and third negative, or last two negative) 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) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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