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Translate the rational 2-torsion point to $(0,0)$ and clear denominators to obtain the integral Weierstrass equation of an elliptic curve. Nonsingularity requires $b\ne 0$ and $b^{\prime }=a^2-4b\ne 0$. The two-isogeny descent usesand its dual isogenyThese formulas extend across their missing affine points as degree-two isogenies of elliptic curves, with kernels $\{O,(0,0)\}$ on their respective curves; substitution gives $\psi \varphi =[2]$ and $\varphi \psi =[2]$.

Define homomorphisms to the square-class group of a field ${\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^2$ byFor ordinary intersections with a line, the product of the three first coordinates is the square of its intercept, proving the homomorphism identity; tangent cases use the same product with multiplicities. For $T=(0,0)$, direct addition gives $x(P+T)=b/x(P)$, so $\alpha (P+T)=[b]\alpha (P)$. Also $\alpha (P)\alpha (-P)=1$ and $\alpha (T{)}^2=1$, which handle inverse points and $T+T=O$. These verify the exceptional values as well. Their kernels are $\psi E^{\prime }(\mathbb{Q})$ and $\varphi E(\mathbb{Q})$. One can check the first assertion directly: if $x=s^2\ne 0$, solving for the first coordinate of a preimage under $\psi$ giveswith rational corresponding ordinate. Conversely, on $E^{\prime }$ one has ${\psi }_x=Y^2/(4X^2)$, a square. The exceptional point $(0,0)$ has a rational preimage exactly when $b$ is a square. The other kernel assertion follows identically, since the double companion curve is isomorphic to $E$ by scaling its coordinates by four and eight.

We now justify the rank factor in the square-class index formula for two-isogeny descent, rather than forgetting a torsion correction. Put $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$, andThe preimage of $\psi \varphi A$ under $\psi$ is $\varphi A+\ker \psi$, soThe Mordell-Weil theorem gives $[A:2A]=2^r\#A[2]$, where $r$ is the rank of an abelian group of $A$. If $b^{\prime }$ is a square, $\#A[2]=4$ and $(0,0)\in \varphi A$, so $\delta =1$. If it is not a square, $\#A[2]=2$ and $\delta =2$. Hence $\delta \#A[2]=4$ in both cases, and

To bound these two images, let $\ell \nmid b$ and take a point with $x\ne 0$. If $v_{\ell }(x)>0$, then $x^2+ax+b$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even. If $v_{\ell }(x)<0$, the leading term $x^3$ uniquely has least valuation in the equation, so $2v_{\ell }(y)=3v_{\ell }(x)$ and again $v_{\ell }(x)$ is even. Thus every image class has a signed square-free integer representative supported on the prime divisors of $b$. The exceptional class $[b]$ also has that property. There are at most $2^{\nu (b)+1}$ such classes. Applying the same reasoning to $E^{\prime }$ gives $\#{\alpha }^{\prime }(B)\le 2^{\nu (b^{\prime })+1}$. The prime-support bound in two-isogeny descent is thereforeHere $\nu$ counts the distinct prime divisors of the absolute value of a nonzero integer; $\nu (1)=\nu (-1)=0$. The nonsingularity hypotheses exclude the otherwise undefined case $\nu (0)$.

For the unheaded practical procedure, enumerate signed square-free divisors $d$ of $b$, and of $b^{\prime }$ on the companion curve. A class $d$ is represented by a point of $E$ exactly when its quartic covering in a two-isogeny descenthas a rational solution with $(U,V)\ne (0,0)$. For $UV\ne 0$, reconstruct $x=d{U}^2/V^2$, $y=dUW/V^3$; substitution verifies the equivalence. Solutions with $V=0$ or $U=0$ account for the identity class or the class of $(0,0)$ respectively. Clearing denominators allows integral $U,V,W$ with $\gcd (U,V)=1$.

Test these finitely many coverings over the real numbers and local fields, especially at two and the primes dividing $b{b}^{\prime }$, to exclude impossible classes. Search the survivors for rational points, close the witnessed classes under multiplication, and use the index formula once both images are determined. Equivalently, local solubility gives a descent upper bound and independent rational points, certified for example by their canonical height of an elliptic curve pairing, give a lower bound. When the bounds coincide the rank is determined. This procedure often succeeds, but local solubility alone does not prove global solubility: a nonzero Tate–Shafarevich group can leave surviving coverings without rational points. In that case the computation gives a rigorous bound rather than a falsely certified exact rank.

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: