Quartic covering in a two-isogeny descent ($C_d$)

For a square class $d$ dividing $b$, the associated quartic covering isIt has a nontrivial rational point exactly when $d$ occurs in the square-class image attached to $E:y^2=x(x^2+ax+b)$.

This quartic covering in a two-isogeny descent has no solution with coprime integers $U,V$ not both zero. If precisely one is odd, the right side is $3$ modulo four. If both are odd, their fourth powers are one modulo sixteen and $14U^2{V}^2=14$ modulo sixteen, so the right side is twelve modulo sixteen. Neither residue is a square. Coprimality excludes the remaining parity case. Consequently the negative unit square class is absent from the two-torsion square-class homomorphism for $Y^2=X^3+14X^2+X$.

For $y^2=x^3+a{x}^2+bx$ with $a\equiv 8$ and $b\equiv 9(\mathrm{mod}16)$, the square class $-1$ in a two-isogeny descent would require coprime integers $u,v$ and an integer $N$ with $N^2=-u^4+a{u}^2{v}^2-b{v}^4$. If $u$ is odd and $v$ even, the right side is $15$ modulo $16$; if $u$ is even and $v$ odd, it is $7$; if both are odd, it is $14$. None is a square modulo $16$. Therefore $-1$ does not occur in the two-torsion square-class homomorphism image.

For $E_5^{\prime }:y^2=x^3+100x$, the two-isogeny descent classes $2$ and $10$ have no rational covering points. Their primitive integral quartics are respectively $N^2=2U^4+50V^4$ and $N^2=10(U^4+V^4)$, with $\gcd (U,V)=1$.

For the first equation, reduction modulo $5$ forces $5\mid U$ and $5\mid N$, since $2$ is not a square modulo $5$ and a nonzero fourth power is $1$. Put $U=5U_1,N=5N_1$. Dividing by $25$ gives $N_1^2=50U_1^4+2V^4$, again a nonsquare modulo $5$ because $5\nmid V$. For the second equation, a primitive pair has $U^4+V^4$ nonzero modulo $5$: it is $1$ or $2$. The right side has five-adic valuation exactly $1$, impossible for a square. Since $(5,25)$ lies on $E_5^{\prime }$, its square-class image is exactly $\{1,5\}$. Combined with the rank bound for a prime congruent-number elliptic curve computation, the two-isogeny rank formula yields rank exactly $1$ for $E_5:y^2=x^3-25x$.

This quartic has no primitive integral solution. If $5\nmid U$, reduction modulo five makes $W^2\equiv 2$, impossible. If $5\mid U$, coprimality gives $5\nmid V$ and $5\mid W$; dividing by twenty-five then gives $(W/5{)}^2\equiv 2V^4\equiv 2(\mathrm{mod}5)$, again impossible. It excludes the square class two in two-isogeny descent on $Y^2=X^3+100X$. The point $(5,25)$ supplies class five, so the subgroup property excludes class ten as well.

The two-isogeny descent quartics associated with square classes $2$ and $-2$ on $Y^2=X^3+24X^2+64X$ have no rational points. A rational solution can be scaled to integral $W,U,V$ with coprime integers $U,V$. If $U$ is odd, reduction modulo sixteen gives $2$ or $10$ for the positive class, and $14$ or $6$ for the negative class; none is a quadratic residue. If $U=2u$ is even, then $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$, with the same sign on the two signed terms. The bracket is odd, so its 2-adic valuation is exactly five, impossible for a nonzero square. These two parity cases exhaust primitive solutions.

If $p\equiv 3(\mathrm{mod}8)$, this quartic covering in a two-isogeny descent has no primitive integral solution. Opposite parities of $U,V$ give $N^2\equiv 2(\mathrm{mod}8)$. If both are odd, their fourth powers are one modulo sixteen and $p^2\equiv 9(\mathrm{mod}16)$, giving $N^2\equiv 20(\mathrm{mod}32)$. Neither residue is a square. This excludes the class $2$ on $Y^2=X^3+4p^{2}X$.

For the curve $Y^2=X^3+4p^{2}X$ with $p\equiv 3(\mathrm{mod}4)$, the classes $p$ and $2p$ in two-isogeny descent lead to $N^2=p(U^4+4V^4)$ and $N^2=2p(U^4+V^4)$. A primitive solution forces $p\mid N$ and $p\nmid V$. The resulting congruences make $-1$ a square modulo $p$, a contradiction. Thus these coverings have no rational points.

A proposed square class in a two-isogeny descent can be excluded by testing its quartic modulo $8$. For $N^2=3U^4+5U^2{V}^2-2V^4$ with $\gcd (U,V)=1$, both variables odd give residue $6$; odd-even gives $3$ or $7$; even-odd gives $6$ or $2$. None is a square modulo $8$, excluding class $3$ and, by the subgroup property of the descent image, its entire coset.