Polynomial Roots Sum 5
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $ P(x) = 2x^3-2ax^2+(a^2-81)x-c $ for some positive integers $ a $ and $ c $. Can you tell me the values of $ a $ and $ c $?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of $ a $." He writes down a positive integer and asks, "Can you tell me the value of $ c $?"
Jon says, "There are still two possible values of $ c $."
Find the sum of the two possible values of $ c $.
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Solution
Let $ r, s, t $ be the three positive integer roots of $ P(x) $. Then by Vieta's formulas, \[\begin{aligned} r+s+t &= a, \\ rs+st+rt &= \frac{a^2-81}{2}, \\ rst &= \frac{c}{2}.\end{aligned}\]Substituting the first equation into the second to eliminate $ a, $ we have \[rs+st+rt = \frac{(r+s+t)^2 - 81}{2} = \frac{(r^2+s^2+t^2) + 2(rs+st+rt) - 81}{2}.\]This simplifies to \[r^2 + s^2 + t^2 = 81.\]Therefore, each of $ r, s, t $ lies in the set $ \{1, 2, \ldots, 9\} $. Assuming without loss of generality that $ r \le s \le t, $ we have $ 81=r^2+s^2+t^2 \le 3t^2, $ so $ t^2 \ge 27, $ and $ t \ge 6 $. We take cases:
If $ t = 6, $ then $ r^2+s^2 = 81 - 6^2 = 45; $ the only solution where $ r \le s \le 6 $ is $ (r, s) = (3, 6) $.
If $ t = 7, $ then $ r^2+s^2 = 81-7^2 = 32; $ the only solution where $ r \le s \le 7 $ is $ (r, s) = (4, 4) $.
If $ t = 8, $ then $ r^2+s^2 = 81-8^2 = 17; $ the only solution where $ r \le s \le 8 $ is $ (r, s) = (1, 4) $.
Therefore, the possible sets of roots of such a polynomial are $ (3, 6, 6), (4, 4, 7), $ and $ (1, 4, 8) $. Calculating $ a = r+s+t $ and $ c=2rst $ for each set, we have $ (a, c) = (15, 216), (15, 224), (13, 64) $.
Since, given the value of $ a, $ there are still two possible values of $ c, $ it must be that $ a = 15, $ since two of the pairs $ (a, c) $ have $ a = 15, $ but only one has $ a = 13 $. Then the sum of the two possible values of $ c $ is \[216 + 224 = \boxed{440}.\]
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