^{1}Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz 53751-71379, Iran.

^{2}Department of Mathematics, Azarbaijan Shahid Madani University, P. O. Box 53751-71379, Tabriz , Iran.

Abstract

Let $E$ be an elliptic curve over $\Bbb{Q}$ with the given Weierstrass equation $ y^2=x^3+ax+b$. If $D$ is a squarefree integer, then let $E^{(D)}$ denote the $D$-quadratic twist of $E$ that is given by $E^{(D)}: y^2=x^3+aD^2x+bD^3$. Let $E^{(D)}(\Bbb{Q})$ be the group of $\Bbb{Q}$-rational points of $E^{(D)}$. It is conjectured by J. Silverman that there are infinitely many primes $p$ for which $E^{(p)}(\Bbb{Q})$ has positive rank, and there are infinitely many primes $q$ for which $E^{(q)}(\Bbb{Q})$ has rank $0$. In this paper, assuming the parity conjecture, we show that for infinitely many primes $p$, the elliptic curve $E_n^{(p)}: y^2=x^3-np^2x$ has odd rank and for infinitely many primes $p$, $E_n^{(p)}(\Bbb{Q})$ has even rank, where $n$ is a positive integer that can be written as biquadrates sums in two different ways, i.e., $n=u^4+v^4=r^4+s^4$, where $u, v, r, s$ are positive integers such that $\gcd(u,v)=\gcd(r,s)=1$. More precisely, we prove that: if $n$ can be written in two different ways as biquartic sums and $p$ is prime, then under the assumption of the parity conjecture $E_n^{(p)}(\Bbb{Q})$ has odd rank (and so a positive rank) as long as $n$ is odd and $p\equiv5, 7\pmod{8}$ or $n$ is even and $p\equiv1\pmod{4}$. In the end, we also compute the ranks of some specific values of $n$ and $p$ explicitly.