1988 USAMO Problems/Problem 5
Let be the polynomial , where are integers. When expanded in powers of , the coefficient of is and the coefficients of , , ..., are all zero. Find .
First, note that if we reverse the order of the coefficients of each factor, then we will obtain a polynomial whose coefficients are exactly the coefficients of in reverse order. Therefore, if we define the polynomial to be noting that if the polynomial has degree , then the coefficient of is , while the coefficients of for are all .
Let be the sum of the th powers of the roots of . In particular, by Vieta's formulas, we know that . Also, by Newton's Sums, as the coefficients of for are all , we find that Thus for . Now we compute . Note that the roots of are all th roots of unity. If , then the sum of nd powers of these roots will be If , then we can multiply by to obtain But as , this is just . Therefore the sum of the nd powers of the roots of is the same as the sum of the nd powers of the roots of The nd power of each of these roots is just , hence the sum of the nd powers of the roots is On the other hand, we can use the same logic to show that Subtracting (2) from (1) and dividing by 32, we find Therefore, .
By a limiting process, we can extend the problem to that of finding a sequence of integers such that (The notation comes from the Alcumus version of this problem.)
If we take logarithmic derivatives on both sides, we get and upon multiplying both sides by , this gives us the somewhat simple form Expanding all the fractions as geometric series, we get Comparing coefficients, we get for all positive integers . In particular, as in Solution 1, we get from which the answer follows.
Remark: To avoid the question of what an infinite product means in the context of formal power series, we could instead view the problem statement as saying that modular arithmetic for polynomials can be defined in exactly the same way as modular arithmetic for integers. Uniqueness of the 's comes from the fact that we have for all by further reduction modulo (as for ), so we could uniquely solve for the 's one at a time. (This idea can be pushed further to explain why it's fine to pass to the infinite product version of the problem.)
To convert the above solution to one that works with polynomials modulo , note that the derivative is not well-defined, as for instance, and are equivalent modulo , but their derivatives, and , are not. However, the operator is well-defined. The other key idea is that for any , modulo , polynomials of the form are invertible, with inverse Therefore, for the polynomial in the problem, call it , we can still form the expression , which is what we originally got by taking the logarithmic derivative and multiplying by , and expand it to eventually get which gets us the same relations (for ).
From the starting point of Solution 2, taking reciprocals and expanding with geometric series gives us On the right, we have the generating function for the number of monic polynomials of degree over the field of two elements, and on the left, we have the factorisation of this generating function that considers the breakdown of any given monic polynomial into monic irreducible factors. As such, we have the interpretation From here, to determine , we analyse the elements of , of which there are in total. Given , if the minimal polynomial of has degree , then and all other roots of appear in . Moreover, if and is an irreducible polynomial of degree , then all roots of appear in . (These statements are all well-known in the theory of finite fields.) As such, for each , there are precisely elements of of degree , and we obtain the same equation as in Solution 2, The rest is as before.
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