From: Jonathan Michaels Claim: Let p be a prime, and let e_k(x_1, x_2, ..., x_(p-1)) be the k-th elementary symmetric polynomial for some k <= p-2. Then e_k(1, 2, ..., p-1) is congruent to 0 (mod p). Proof: Consider the following function: (1) f(x) = (x-1)*(x-2)*...*(x-(p-1)). It consists of the product of p-1 terms, each of which is a polynomial of degree one, so f(x) is a polynomial of degree p-1. Hence, f(x) can also be written in standard polynomial form as follows: (2) f(x) = a_0*x^(p-1) + a_1*x^(p-2) + ... + a_(p-1) where all a_k are as-yet undetermined constants. Note that f(x) expressed as a product of p-1 single-degree polynomials as in (1) can be expanded using repeated applications of the distributive law of multiplication over addition, resulting in 2^(p-1) terms. This is because each term is the product of (either x or -1)*(either x or -2)*...*(either x or -(p-1)), that is, the product of p-1 factors, for each of which there are 2 independent choices. Thus by the Product Rule, 2^(p-1) terms result. Lemma 0.1: In (2) above, a_0 = 1. Only one of the abovementioned 2^(p-1) terms contributes to the x^(p-1) term of polynomial (2), because to have a power of x to the p-1, each of the above choices must be 'x', which can happen in only one way. Furthermore, since all of the factors in the term that contribute to the x^(p-1) term of polynomial (2) are x, none of the factors are constants, so the coefficient of said term, a_0, is 1. Lemma 0.2: In (2) above, a_(p-1) is congruent to -1 (mod p). Again, only one of the abovementioned 2^(p-1) terms contribute to the constant term of the polynomial (2), because each of the choices of factors must /not/ be 'x', that is, they must choose the constant. Thus, the constant term of the polynomial (2), a_(p-1), is (-1)*(-2)*...*(-(p-1)) = (-1)^(p-1) * (p-1)!. (-1)^(p-1) is congruent to 1 by Fermat's Little Theorem, and (p-1)! is congruent to -1 by Wilson's Theorem (I can prove this quickly if you want.) Thus, a_(p-1) is congruent to 1*(-1) = -1 (mod p). Lemma 0.3: In (2) above, a_k = (-1)^k * e_k(1, 2, ..., p-1) for all k such that 0 < k < p-1. To see this, we must look closely at the expansion of (1) into 2^(p-1) terms. The different terms that contribute to the x^(p-k-1) term of (2) consist of all possible ways to multiply together p-k-1 'x's and k constants {-1, -2, ..., -(p-1)} from the p-1 terms of (1), and these different terms are added up to yield the coefficient a_k of the x^(p-k-1) term of (2). Put another way, a_k is equal to the sum of all the possible products of k unique factors chosen from {-1, -2, ..., -(-p-1)}. Note that this is precisely the definition of e_k(-1, -2, ..., -(p-1)) where e_k is the k-th elementary symmetric polynomial. Furthermore, from each product term of the sum that forms a_k, a (-1)^k can be factored out (leaving all positive factors), and since this (-1)^k is present in each term of the sum, it can be removed from the whole thing, showing that a_k = (-1)^k * e_k(1, 2, ..., p-1). Now we are ready to prove the main result. Note that for all x in {1, 2, ..., p-1}, f(x) = 0 because one of the terms in (1) will be equal to zero. So, using the form of (2): (3) a_0*x^(p-1) + a_1*x^(p-2) + ... + a_(p-1) = 0 for all x in {1, 2, ..., p-1}. Now we work mod p. x^(p-1) is congruent to 1 mod p by Fermat's Little Theorem, and a_0 = 1 by Lemma 0.1, so the first term on the left side of (3) is congruent to 1*1 = 1. The last term, a_(p-1), is congruent to -1 mod p by Lemma 0.2, which gives: (4) 1 + a_1*x^(p-2) + a_2*x^(p-3) + ... + a_(p-2)*x - 1 = 0 (mod p) for all x in {1, 2, ..., p-1}. Cancelling the 1 and -1, (5) a_1*x^(p-2) + a_2*x^(p-3) + ... + a_(p-2)*x = 0 (mod p) for all x in {1, 2, ..., p-1}. Now, note that we have an unknown modular polynomial of degree p-2, and we know its value at p-1 points. We have enough points to uniquely determine the polynomial; that is, there exists a unique set of values a_1, a_2, ..., a_(p-2) mod p that satisfy (5). The set of values a_1 = a_2 = ... = a_(p-2) = 0 (mod p) satisfies (5), and by uniqueness, this is the only such set. This brings us to our conclusion. a_k = 0 (mod p) for all k such that 0 < k < p-1, and from Lemma 0.3, a_k = (-1)^k * e_k(1, 2, ..., p-1) for all k such that 0 < k < p-1; thus: (6) (-1)^k * e_k(1, 2, ..., p-1) = 0 (mod p) for all k such that 0 < k < p-1, and since (-1)^k is never equal to 0, it must be the case that e_k(1, 2, ..., p-1) = 0 (mod p) for all k such that 0 < k < p-1. QED. Jonathan