Modulo-thirty-two obstruction for a congruent-number descent class ($N^2=2(U^4+p^2{V}^4)$)

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 $Y^2=X^3+4p^{2}X$.