The Somos 5 sequences are a family of sequences defined by a fifth order bilinear recurrence relation with constant coefficients. For particular choices of coefficients and initial data, integer sequences arise. By making the connection with a second order nonlinear mapping with a first integral, we prove that the two subsequences of odd/even index terms each satisfy a Somos 4 (fourth order) recurrence. This leads directly to the explicit solution of the initial value problem for the Somos 5 sequences in terms of the Weierstrass sigma function for an associated elliptic curve.

In this paper we present a proof that there exist innitely many rational sided triangles with two rational medians and rational area. These triangles correspond to rational points on an elliptic curve of rank one. We also display three triangles (one previously unpublished) which do not belong to any of the known innite families. I.

Original article is available at

Abstract. In his paper Triangles with three rational medians, about the characterization of all rational-sided triangles with three rational medians, Buchholz proves that each such triangle corresponds to a point on a oneparameter family of elliptic curves whose rank is at least 2. We prove that in fact the exact rank of the family in Buchholz paper is 3. We also exhibit a subfamily whose rank is at least 4 and we prove the existence of in nitely many curves of rank 5 over Q parametrized by an elliptic curve of positive rank. Finally, we show particular examples of curves within those families having rank 9 and 10 over Q. 1.

Triangles having rational sides a, b, c and rational area Q are called Heron triangles. Associated to each Heron triangle is the quartic v 2 = u(u − a)(u − b)(u − c). The Heron formula states that Q = √ P (P − a)(P − b)(P − c) where P is the semi-perimeter of the triangle, so the point (u, v) = (P, Q) is a rational point on the quartic. Also the point of in nity is on the quartic. By a standard construction it can be proved that the quartic is equivalent to the elliptic curve y 2 = (x + a b)(x + b c)(x + c a). The point (P, Q) on the quartic transforms to −2abc (x, y) =