= Signed divisor-two obstruction for an isogeny covering
{title2=$W^2=\pm2U^4+24U^2V^2\pm32V^4$}
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 $32(\pm u^4+3u^2V^2\pm V^4)$, 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.
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