2008 iTest Problems/Problem 84
Contents
Problem
Let be the sum of all integers for which the polynomial can be factored over the integers. Compute .
Solutions
Solution 1 (credit to official solution)
Let the roots of the quadratic be and . By Vieta's Formulas, and = .
We know that one of the possible values of is 0 because has integer roots. However, adding or removing 0 does not affect the value of , so we can divide both sides by . Doing so results in
WLOG, let be a factor of , so and . Thus,
Since can be positive or negative, the positive values cancel with the negative values. The prime factorization of is , so there are positive factors that are less than . Thus, there are a total of values of , so the absolute value of the sum of all values of equals .
Solution 2
The discriminant of the function is . Since all roots are integers and leading term is 1, the discriminant must equal , where is an integer.
Thus, we know that
Let be a factor of , so
Note that if is odd, then is even, so can not be an integer. Thus, must be even. Let , so .
If , then . Also, since can be positive or negative, the positive values cancel with the negative values. So WLOG, let .
The prime factorization of is , so there are positive factors that are less than . Thus, there are a total of values of , so the absolute value of the sum of all values of equals .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 83 |
Followed by: Problem 85 | |
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