Question

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 $c.$

Answer

Answer: 440

Concepts:Vieta’s Formulas polynomial logical deduction casework

Difficulty rating: 2500

Solution:

Dividing by $2,$ the roots $r \le s \le t$ satisfy $r + s + t = a,$ $rs + rt + st = \frac{a^2 - 81}{2},$ and $rst = \frac{c}{2}.$ Therefore $\begin{aligned} &r^2 + s^2 + t^2 \\ &= (r + s + t)^2 - 2(rs + rt + st) \\ &= a^2 - (a^2 - 81) \\ &= 81. \end{aligned}$

The triples of positive integers whose squares sum to $81$ are $(1, 4, 8),$ $(4, 4, 7),$ and $(3, 6, 6),$ with $a = r + s + t$ equal to $13,$ $15,$ and $15.$ Since knowing $a$ still left Jon two choices, $a = 15,$ and the two polynomials come from $(4, 4, 7)$ and $(3, 6, 6).$

The corresponding values of $c = 2rst$ are $2 \cdot 4 \cdot 4 \cdot 7 = 224$ and $2 \cdot 3 \cdot 6 \cdot 6 = 216,$ with sum $224 + 216 = 440.$

2015 AIME II Problem 6

Answer: 440

Difficulty rating: 2500

Solution:

Problem 6 in Other Years