[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:

AIME_I_2013_10_01

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:

AIME_I_2013_10_02

(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:

AIME_I_2013_10_03

(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!

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