2013 AMC 12B Problems/Problem 25
Problem
Let be the set of polynomials of the form where are integers and has distinct roots of the form with and integers. How many polynomials are in ?
Solution 1
If we factor into irreducible polynomials (in ), each factor has exponent in the factorization and degree at most (since the with come in conjugate pairs with product ). Clearly we want the product of constant terms of these polynomials to equal ; for , let be the number of permitted with constant term . It's easy to compute , , , , , , and obviously for negative .
Note that by the distinctness condition, the only constant terms that can be repeated are those with and , i.e. and . Also, the s don't affect the product, so we can simply count the number of polynomials with no constant terms of and multiply by at the end.
We do casework on the (unique) even constant term in our product. For convenience, let be the number of ways to get a product of without using (so only using ) and recall ; then our final answer will be . It's easy to compute , , , , , , so we get
Solution 2
Disregard sign; we can tack on if the product ends up being negative.
(2) (1 is not included)
(4)
(6)
(6)
(7)
(8)
Our answer is
Solution 3
By Vieta's formula is the product of all roots. As the roots are all in the form , there must exist a conjugate for each root.
If , the roots can be , , , , totaling pairs of roots.
If , the roots can be , , totaling pairs of roots.
If , , the roots can be , , totaling pairs of roots.
For each case can be added, yielding 2 more cases .
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
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Followed by Last Question |
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