2015 AMC 12A Problems/Problem 18
Contents
Problem
The zeros of the function are integers. What is the sum of the possible values of ?
Solution 1
The problem asks us to find the sum of every integer value of such that the roots of are both integers.
The quadratic formula gives the roots of the quadratic equation:
As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant equals , for some nonnegative integer .
From this last equation, we are given a hint of the Pythagorean theorem. Thus, must be a Pythagorean triple unless .
In the case , the equation simplifies to . From this equation, we have . For both and , yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")
If is a positive integer, then only one Pythagorean triple could match the triple because the only Pythagorean triple with a as one of the values is the classic triple. Here, and . Hence, . Again, yields two integers for both and , so these two values also satisfy the original constraints.
There are a total of four possible values for : and . Hence, the sum of all of the possible values of is .
Solution 2
By the quadratic formula, the roots can be represented by For , , since and will have different mantissas (mantissae?).
Now observe the discriminant and have two cases.
Positive
and , since gives imaginary roots. Testing positive values, quickly see that . After and , the difference between the closest nonzero factor pairs of perfect squares exceeds . For , . Checking both yields an integer.
Negative
We can instead test with . If , we have our original expression. Thus, for the same reasons, . (0 does not affect the answer).
(Solution by BJHHar)
Solution 3
Let and be the roots of
By Vieta's Formulas, and
Substituting gets us
Using Simon's Favorite Factoring Trick:
This means that the values for are giving us values of and . Adding these up gets .
Solution 4
The quadratic formula gives . For to be an integer, it is necessary (and sufficient!) that to be a perfect square. So we have ; this is a quadratic in itself and the quadratic formula gives
We want to be a perfect square. From smartly trying small values of , we find as solutions, which correspond to . These are the only ones; if we want to make sure then we must hand check up to . Indeed, for we have that the differences between consecutive squares are greater than so we can't have be a perfect square. So summing our values for we find . as the answer.
Additional note: You can use the quadratic and plug in squares for a (since for to be an integer would have to be some square), and eventually you can notice a limit to get the answer~
Solution 5
First of all, we know that is the sum of the quadratic's two roots, by Vieta's formulas. Thus, must be an integer. Then, we notice that the discriminant must be equal to a perfect square so that the roots are integers. Thus, where is an integer.
We can complete the square and rearrange to get . Let's define , just to make things a little easier to write, so now we have . We can now list out the integer factor pairs of 16 and the resulting values of and . (Note that and must both be integers)
Doesn't work
Doesn't work
Doesn't work
Doesn't work
We want the possible values of , which are and . As , can equal or Adding all of that up gets us our answer, .
(Solution by Curious_crow)
Solution 6 (big fast)
Let our roots be and . By vieta's formulas, we have that and . Thus, , or .
4 can only be obtained from (2,2), (1,4), (-2,-2), or (-1,-4), giving us and , which add up to
-skibbysiggy
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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