On Silverman's conjecture for a family of elliptic curves

Let $E$ be an elliptic curve over $\mathbb{Q}$ 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 $E^{(D)}: y^2=x^3+aD^2x+bD^3$\mathrm{.}Let$E^{(D)}(\Bbb{Q})$bethegroupof$\Bbb{Q}$-rational points of $E^{(D)}$. It is conjectured by J. Silverman that there are infinitely many primes $p$ for which $E^{(p)}(\mathbb{Q})$ has positive rank, and there are infinitely many primes $q$ for which $E^{(q)}(\mathbb{Q})$ has rank $0$. In this paper, assuming the parity conjecture, we show that for infinitely many primes $p$, the elliptic curve $E_n^{(p)}:y^2=x^3-n{p}^{2}x$ has odd rank and for infinitely many primes $p$, $E_n^{(p)}(\mathbb{Q})$ 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 $E_n^{(p)}(\mathbb{Q})$ has odd rank (and so a positive rank) as long as $n$ is odd and $p\equiv 5,7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ or $n$ is even and $p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$. In the end, we also compute the ranks of some specific values of $n$ and $p$ explicitly.