= Solution
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 $y\ne0$ on
$$
E_D:y^2=x^3-D^2x,
$$
the formulas
$$
a=\frac{x^2-D^2}{y},
\qquad b=\frac{2Dx}{y},
\qquad c=\frac{x^2+D^2}{y}
$$
give $a^2+b^2=c^2$ and $ab/2=D$, after changing signs if necessary. Conversely, a rational right triangle of area $D$ gives
$$
x=\frac{D(a+c)}b,
\qquad y=\frac{2D^2(a+c)}{b^2},
$$
so 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 $p\equiv3\pmod4$ for which $p\not\equiv-1\pmod\ell$. 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 $p\equiv3\pmod8$ shows that the rational $2$-primary torsion has order at most four. Since
$$
O,(0,0),(D,0),(-D,0)
$$
already form the full rational <2-torsion>,
$$
E_D(\mathbb Q)_{\mathrm{tors}}\cong(\mathbb Z/2\mathbb Z)^2.
$$
The triangle construction uses exactly the points with $y\ne0$, which are therefore nontorsion. By the <Mordell-Weil theorem>, such a point exists exactly when the free part has positive rank. Hence
$$
\boxed{D\text{ is congruent}\quad\Longleftrightarrow\quad\operatorname{rank}E_D(\mathbb Q)\geq1.}
$$
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