**2013 AIME I 10.** There are nonzero integers , , , and such that the complex number is a zero of the polynomial . For each possible combination of and , let be the sum of the zeroes of .Find the sum of the ‘s for all possible combinations of and .

* For polynomials with real coefficients, complex roots come in conjugate pairs.* Since we know that is a root of , it follows that is also a root of . It also means that the third root of must be real. Let be this third root.

**We can apply Vieta’s formulas to and obtain 3 equations in , and :**

This simplifies to

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

- Since and are integers, implies that
**is also an integer.** - Now, we can use to say that
(Note that we need the previous point above to establish this, since we only knew that was real, but we didn’t know that it had to be integral.)**is a factor of 65.** - Looking further, since , implies further that
. Hence, we have .**is a positive factor of 65** - can’t be 65: if it is, implies that , which means that one of and must be zero (but the problem forbids this).

We are left with 3 possibilities for 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:* .

The 3 equations simplify to

Notice that the value of is not important, in that whether , the values of that result are the same. Hence, * we only have to focus on the possible values of .* (This applies across the other cases as well.)

We have the following solutions in this case:

*Case 2:* .

The 3 equations simplify to

We have the following solutions in this case:

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

*Case 3:* .

The 3 equations simplify to

We have the following solutions in this case:

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

Now we sum up all the possible values of :

**Done!**