[Soln] 2015 AIME II Problem 6

2015 AIME II 6. 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.

With low-degree polynomial questions where roots are mentioned, it should almost be second-nature to label the roots (say p, q, r) and apply Vieta’s formulas. For this polynomial P, we obtain the equations

\begin{aligned} p + q + r &= a &&-(1) \\  pq + qr + rp &= \displaystyle\frac{a^2 - 81}{2} &&-(2) \\  pqr &= \displaystyle\frac{c}{2} &&-(3) \end{aligned}

With equations like (1) and (2), it should also be second nature to do the following to obtain the sum of squares of the roots:

\begin{aligned} p^2 + q^2 + r^2 &= (p+q+r)^2 - 2(pq + qr + rp) \\  &= a^2 - (a^2 - 81), \\  p^2 + q^2 + r^2 &= 81. \hspace{1em} -(4) \end{aligned}

That cancelled nicely! Since we know that p, q and r are positive integers, (4) tells us that there can’t be too many possibilities for \{ p, q, r \}. Indeed, we can list out all the possibilities systematically:

\begin{aligned} \{ p, q, r \} &= \{ 8, 4, 1 \} \text{ or } \\  &= \{ 7, 4, 4 \} \text{ or } \\  &= \{ 6, 6, 3 \}.\end{aligned}

Let’s take stock here. Given that the polynomial P has the form given in the problem and that its roots are all positive integers, we have determined that there are 3 possibilities for P, namely those which have the roots \{ 8, 4, 1 \}, \{ 7, 4, 4\} or \{ 6, 6, 3 \}. Now, let’s see how Steve & Jon’s conversation can help us narrow down the possibilities.

From the conversation, we know that knowing the value of a cannot tell us what P is. Using equation (1), let’s see what values of a are possible in the first place:

\begin{aligned} &\underline{Roots of } P &&\underline{Value of } a \\  &8, 4, 1 &&13 \\  &7, 4, 4 &&15 \\  &6, 6, 3 &&15 \end{aligned}

Well, this means that P can’t have roots \{ 8, 4, 1 \}: if P had those roots, then knowing the value of a would reveal P to us!

Hence, we must have a = 15. Equation (3) gives us the 2 possible values of c:

c = 2 (7)(4)(4) = 224 \hspace{1em} or \hspace{1em} c = 2(6)(6)(3) = 216.

Hence, the desired answer is 224 + 216 = 440. Done.

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