2012 AMC 12B Problems/Problem 23
Problem 23
Consider all polynomials of a complex variable, , where and are integers, , and the polynomial has a zero with What is the sum of all values over all the polynomials with these properties?
Solution (doubtful)
Since is a root of , and has integer coefficients, must be algebraic. Since is algebraic and lies on the unit circle, must be a root of unity (Comment: this is not true. See this link: [1]). Since has degree 4, it seems reasonable (and we will assume this only temporarily) that must be a 2nd, 3rd, or 4th root of unity. These are among the set . Since complex roots of polynomials come in conjugate pairs, we have that has one (or more) of the following factors: , , , or . If then ; a contradiction since are non-negative. On the other hand, suppose . Then . This implies while correspondingly. After listing cases, the only such valid are , , , , and .
Now suppose . Then whereupon and . But then and . This gives only the cases equals , which we have already counted in a previous case.
Suppose . Then so that and . This only gives rise to equal which we have previously counted.
Finally suppose divides . Using polynomial division ((or that to make the same deductions) we ultimately obtain that . This can only happen if is .
Hence we've the polynomials However, by inspection has roots on the unit circle, because which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that in an th root of unity where , and is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If is prime, then \textit{every} th root of unity except 1 must satisfy our polynomial, but since and the degree of our polynomial is 4, this is impossible. Suppose is composite. If it has a prime factor greater than 5 then again every th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose is divisible only by 2,3,or 5. Since by hypothesis is not a 2nd or 3rd root of unity, must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy . But has exactly all 5th roots of unity excluding 1, and . Thus this must divide which implies . This completes the proof.
Solution
First, assume that , so or . does not work because . Assume that . Then , we have , so . Also, has to be true since . Now gives , therefore the only possible choices for are . In these cases, . The sum of over these cases is .
Second, assume that , so for some real , . By conjugate roots theorem we have that , therefore is a factor of , and we may assume that
for some real . Expanding this polynomial and comparing the coefficients, we have the following equations:
From the first and the third we may deduce that and that , if (we will consider by the end). Let . From the second equation, we know that is non-negative.
Consider the following cases:
Case 1: . Then , , so , . However, this has already been found (i.e. the form of ).
Case 2: . Then since , we have . However, , therefore . This is true only when . Also, we get again. In this case, , so , , . has a root . .
Last case: . We have and that has a root . .
Therefore the desired sum is .
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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