Diophantine Equation $X^4+Y^4=2(U^4+V^4)$
In this paper, the theory of elliptic curves is used for finding the solutions of the quartic Diophantine equation $X^4+Y^4=2(U^4+V^4)$ Keywords: Diophantine equation, Elliptic curve, Congruent number
arXiv subjects
Publications and source records attributed to Kamran Nabardi.
In this paper, the theory of elliptic curves is used for finding the solutions of the quartic Diophantine equation $X^4+Y^4=2(U^4+V^4)$ Keywords: Diophantine equation, Elliptic curve, Congruent number
In this paper, we construct a family of elliptic curves with rank $\geq 5$. To do this, we use the Heron formula for a triple $(A^2, B^2, C^2)$ which are not necessarily the three sides of a triangle. It turns out that as parameters of a family of elliptic curves, these three positive integers $A$, $B$, and $C$, along with the extra parameter $D$ satisfy the quartic Diophantine equation $A^4+D^4=2(B^4+D^4)$.
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.