## [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.