2015 AIME II Problems/Problem 6
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form for some positive integers and . Can you tell me the values of and ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of ." He writes down a positive integer and asks, "Can you tell me the value of ?"
Jon says, "There are still two possible values of ."
Find the sum of the two possible values of .
Solution 1 (Algebra)
We call the three roots (some may be equal to one another) , , and . Using Vieta's formulas, we get , , and .
Squaring our first equation we get .
We can then subtract twice our second equation to get .
Simplifying the right side:
So, we know .
We can then list out all the triples of positive integers whose squares sum to :
We get , , and .
These triples give values of , , and , respectively, and values of , , and , respectively.
We know that Jon still found two possible values of when Steve told him the value, so the value must be . Thus, the two values are and , which sum to .
Solution 2 (Algebra+ Brute Force)
First things first. Vietas gives us the following:
From (2), must have odd parity, meaning must be a multiple of 4, which implies that both sides of (2) are even. Then, from (1), we see that an odd number of , , and must be odd, because we have already deduced that is odd. In order for both sides of (2) to be even, there must only be one odd number and 2 even numbers.
Now, the theoretical maximum value of the left side of (2) is . That means that the maximum bound of is where
which simplifies to
So now we have that from (2), from (5), and is odd from (2). This means that could equal , , or . At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of , , and , of which the last two return equal values. Then, AWD.
Thank you, Rowechen Zhong.
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