= Five-adic obstructions for the congruent-number isogeny covers
{title2=$\alpha'(E'_5)=\{1,5\}$}
For $E'_5:y^2=x^3+100x$, 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 $E'_5$, 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 $E_5:y^2=x^3-25x$.
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