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If $p\mid D$, square-freeness gives the cuspidal reduction $y^2=x^3$. Its nonsingular group is isomorphic to the additive group ${\mathbb{F}}_p$, so it is cyclic of order $p$.

Suppose $p\nmid D$. The reduction is an elliptic curve. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is a quadratic nonresidue. Pairing $x$ with $-x$ in the quadratic-character sum and usingshows that the two contributions cancel. Therefore $\#\tilde{E}({\mathbb{F}}_p)=p+1$. The three roots $0,D,-D$ are distinct, so the full 2-torsion is rational. A cyclic group has at most two elements killed by $2$, hence $\tilde{E}({\mathbb{F}}_p)$ is noncyclic. Thus

A positive integer $D$ is a congruent number when it is the area of a right triangle with positive rational side lengths. For a point $P=(x,y)$ with $y\ne 0$ onthe formulasgive $a^2+b^2=c^2$ and $ab/2=D$, after changing signs if necessary. Conversely, a rational right triangle of area $D$ givesso these constructions are inverse up to the usual sign choices.

It remains to distinguish torsion. If an odd prime $\ell$ divided the order of a rational torsion point, choose by the Dirichlet theorem on primes in arithmetic progressions a good prime $p\equiv 3(\mathrm{mod}4)$ for which $p\not\equiv -1(\mathrm{mod}\ell )$. Part (a) gives $\#E_D({\mathbb{F}}_p)=p+1$, while part (b), applied to the formal group of an elliptic curve, makes reduction injective on $\ell$-power torsion because $\ell$ is a unit in ${\mathbb{Z}}_p$. This is impossible. Similarly, a good prime $p\equiv 3(\mathrm{mod}8)$ shows that the rational $2$-primary torsion has order at most four. Sincealready form the full rational 2-torsion,The triangle construction uses exactly the points with $y\ne 0$, which are therefore nontorsion. By the Mordell-Weil theorem, such a point exists exactly when the free part has positive rank. Hence

Use the two-descent on an elliptic curve map associated with the three rational roots $0,-1,-4$,with the standard limiting values at the 2-torsion. Only the square classes of $-1$, $2$ and $3$ can occur, because all other numerator and denominator valuations in the three factors $x,x+1,x+4$ are even. Checking solubility over $\mathbb{R}$, ${\mathbb{Q}}_2$ and ${\mathbb{Q}}_3$ leaves exactlyThese four classes are represented respectively by $O$, $P_1$, $P_2$ and $P_1+P_2$. Thus $|E(\mathbb{Q})/2E(\mathbb{Q})|=4$. Since part (b) givesthe quotient by doubling has order $2^{r+2}$. Therefore

When an elliptic curve has full rational 2-torsion, its Kummer map embeds $E(\mathbb{Q})/2E(\mathbb{Q})$ into a product of square-class groups. Valuation and local-solubility conditions reduce the image to finitely many classes and thereby bound the rank of the Mordell-Weil group.