 Negative unit obstruction for the fourteen-coefficient isogeny cover

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 $Y^2=X^3+14X^2+X$.