Colloquium| Institute of Mathematical Sciences
Time：10:30-11:30, Nov. 29, Friday
Location：Room S408, IMS
Speaker: Hourong Qin , Nanjing University
Abstract：We show that if a square-free and odd (respectively, even) positive integer n is a congruent number, then #{(x,y,z) ∈Z3|n = x2+2y2+32z2} = #{(x,y,z) ∈Z3|n = 2x2+4y2+9z2−4yz}, respectively, #{(x,y,z) ∈Z3| n 2 = x2+4y2+32z2} = #{(x,y,z) ∈Z3| n 2 = 4x2+4y2+9z2−4yz}. If we assume that the weak Brich-Swinnerton-Dyer conjecture is true for the elliptic curves En : y2 = x3 −n2x, then, conversely, these equalities imply that n is a congruent number. We shall also discuss some applications of the proposed method. In particular, for a prime p, we show that if p ≡ 1 (mod 8) is a congruent number, then the 8-rank of K2OQ(√p) equals one, and if p ≡ 1 (mod 16) with h(−p) 6≡ h(−2p) (mod 16), then 2p is not a congruent number.