By Knopp M.I. (ed.)

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Hence ϕV (S) = n if V = 1 and ϕV (S) = q + 1 otherwise. Finally, let V be the stabilizer of two letters. Then ϕV (S) = 2, since V has order q − 1 = 2. Since 0 ≡ q 3 + 1 ≡ 2 ≡ q + 1 ≡ n mod 2, the even-odd trick does not reduce the number of roots. 3). Recently, McGarraghy found some other examples which show that the LewisMcGarraghy-polynomial improves the Beaulieu-Palfrey result. Let us give another example. 5. Let G = AGL(n, q), n ≥ 2 be the affine linear group over Fq . Consider the action of G on the vector space V = Fnq by semilinear transformations and let H = G0 be the stabilizer of the zero vector.

Last year, in one of our semi-regular conversations I tempted Manjul Bhargava into trying his hand at the difficult job of proving the 290-conjecture. Manjul started the task by reproving the 15-theorem, and now he has discovered the particularly simple proof he gives in the following paper, which has made it unnecessary for us to publish our rather more complicated proof. Manjul has also proved the “33-theorem” – much more difficult than the 15-theorem – which asserts that an integer-matrix form will represent all odd numbers provided only that it represents 1, 3, 5, 7, 11, 15, and 33.

Sloane, Sphere packings, lattices and groups, Grundlehren der Mathematischen Wissenschaften 290, Springer-Verlag, New York, 1999. edu Contemporary Mathematics On the Conway-Schneeberger Fifteen Theorem Manjul Bhargava Abstract. This paper gives a proof of the Conway-Schneeberger Fifteen Theorem on the representation of integers by quadratic forms, to which the paper of Conway [1] in these proceedings is an extended foreword. 1. Introduction. In 1993, Conway and Schneeberger announced the following remarkable result: Theorem 1 (“The Fifteen Theorem”).