## [Soln] 2013 AIME I Problem 10

2013 AIME I 10. There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r + si$ is a zero of the polynomial $P(x) = x^3 - ax^2 + bx - 65$. For each possible combination of $a$ and $b$, let $p_{a,b}$ be the sum of the zeroes of $P(x)$.Find the sum of the $p_{a,b}$‘s for all possible combinations of $a$ and $b$.

For polynomials with real coefficients, complex roots come in conjugate pairs. Since we know that $r + si$ is a root of $P$, it follows that $r - si$ is also a root of $P$. It also means that the third root of $P$ must be real. Let $\alpha$ be this third root.

We can apply Vieta’s formulas to $P$ and obtain 3 equations in $r$, $s$ and $\alpha$:

$\begin{cases} \alpha + (r+si) + (r-si) &= a, \\ \alpha(r+si) + \alpha(r-si) + (r+si)(r-si) &= b, \\ \alpha (r+si)(r-si) &= 65. \end{cases}$

This simplifies to

\begin{aligned} \alpha + 2r &= a &&-(1) \\ 2\alpha r + r^2 + s^2 &= b &&-(2) \\ \alpha(r^2 + s^2) &= 65 &&-(3) \end{aligned}

(Note that $(3)$ can be substituted directly into $(2)$. We won’t do it right now but let’s keep it in mind.) Note also that Vieta’s formulas show us that $p_{a,b} = a$, so the problem is equivalent to finding the sum of all $a$ for valid combinations of $(a, b)$.

• Since $a$ and $r$ are integers, $(1)$ implies that $\alpha$ is also an integer.
• Now, we can use $(3)$ to say that $\alpha$ is a factor of 65. (Note that we need the previous point above to establish this, since we only knew that $\alpha$ was real, but we didn’t know that it had to be integral.)
• Looking further, since $r^2 + s^2 > 0$, $(3)$ implies further that $\alpha$ is a positive factor of 65. Hence, we have $\alpha = 1, 5, 13, \text{ or } 65$.
• $\alpha$ can’t be 65: if it is, $(3)$ implies that $r^2 + s^2 = 1$, which means that one of $r$ and $s$ must be zero (but the problem forbids this).

We are left with 3 possibilities for $\alpha$ which we will go through in turn. This is straightforward case checking; the tricky part is in making sure we enumerate ALL possibilities.

Case 1: $\alpha = 1$.

The 3 equations simplify to

\begin{aligned} &a &&= 2r + 1, \\ &b &&= 2r + 65, \\ &r^2 + s^2 &&= 65. \end{aligned}

Notice that the value of $s$ is not important, in that whether $s = \beta \text{ or } -\beta$, the values of $(a,b)$ that result are the same. Hence, we only have to focus on the possible values of $r$. (This applies across the other cases as well.)

We have the following solutions in this case:

Case 2: $\alpha = 5$.

The 3 equations simplify to

\begin{aligned} &a &&= 2r + 5, \\ &b &&= 10r + 65, \\ &r^2 + s^2 &&= 13. \end{aligned}

We have the following solutions in this case:

(A quick check shows that the combinations $(a,b)$ here don’t overlap with the previous case.)

Case 3: $\alpha = 13$.

The 3 equations simplify to

\begin{aligned} &a &&= 2r + 13, \\ &b &&= 26r + 65, \\ &r^2 + s^2 &&= 5. \end{aligned}

We have the following solutions in this case:

(Again, it’s easy to do a quick check to see that none of the $(a,b)$ are repeated in the other cases.)

Now we sum up all the possible values of $p_{a,b} = a$:

\begin{aligned} \text{sum} &= (-15 -13 -7 -1 +3 +9 +15 +17) \\ &\hspace{2em} + (-1 + 1 + 9 + 11) + (9+ 11+15+17) \\ &= 8 + 20 + 52 \\ &= \fbox{80}. \end{aligned}

Done!