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Take the odd congruence class $p\equiv 3(\mathrm{mod}8)$. The congruent number elliptic curve in the standard scaled coordinates isThese are the two curves in two-isogeny descent. On $E_p$, the prime-support bound in two-isogeny descent restricts the first image to $\{1,-1,p,-p\}$. All four classes occur: the 2-torsion points give $\alpha ((0,0))=[-p^2]=[-1]$, $\alpha ((p,0))=[p]$, and $\alpha ((-p,0))=[-p]$. Thus $\#\alpha (E_p)=4$.

On $E_p^{\prime }$, the equation forces $x>0$ whenever $x\ne 0$. Its exceptional value is $[4p^2]=1$. The second image is therefore contained in $\{1,2,p,2p\}$. Use the quartic covering in a two-isogeny descent to exclude its three nontrivial candidates.

For $d=p$, a rational solution can be written with coprime integral $U,V$, and must satisfyThen $p\mid W$, so $U^4+(2V^2{)}^2\equiv 0(\mathrm{mod}p)$. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is not a quadratic residue; a sum of two squares is zero only if both are zero. This forces $p\mid U,V$, contradicting coprimality. For $d=2p$, the same reasoning applies to $W^2=2p(U^4+V^4)$ and again forces $p\mid U,V$.

For $d=2$, the equation isIt implies $W=2Z$, whence $2Z^2=U^4+p^2{V}^4$. Exactly one of $U,V$ odd is impossible by parity, while both even contradict coprimality. Thus both are odd. Since $p\equiv 3(\mathrm{mod}8)$ implies $p^2\equiv 9(\mathrm{mod}16)$, reduction modulo sixteen gives $2Z^2\equiv 1+9=10$. But the possible residues of $2Z^2$ modulo sixteen are $0,2,8$. This contradiction eliminates $d=2$. Consequently ${\alpha }^{\prime }(E_p^{\prime })=\{1\}$ and the square-class index formula for two-isogeny descent givesThe reduction argument for $y^2=x^3+kx$ used earlier bounds rational torsion by four for every nonzero integer $k$: its point-count cancellation at $q\equiv 3(\mathrm{mod}4)$ does not depend on the sign of $k$, and the same prime choices apply. Here $k=-p^2$ and all four 2-torsion points are rational. Hence $E_p(\mathbb{Q})$ consists exactly of $O,(0,0),(p,0),(-p,0)$.

To relate this calculation to congruent numbers, a rational point $(x,y)$ with $y\ne 0$ produces the right triangle with side lengthsThe identities $A^2+B^2=C^2$ and $AB/2=p$ follow from $y^2=x(x^2-p^2)$. Conversely, a positive rational right triangle of area $p$ with legs $A,B$ and hypotenuse $C$ givesDirect substitution uses $AB=2p$ and $C^2=A^2+B^2$. No such point exists here. This proves the noncongruent primes congruent to three modulo eight: