Congruent number problem

Remarks

The problem reduces to square-free $n$, since $n$ is congruent if and only if $nk^2$ is congruent for every positive integer $k$. The number $6$ is congruent, by the triangle with sides $3, 4, 5$, which is a Pythagorean triple. The number $5$ is congruent: the sides $3/2$, $20/3$, $41/6$ satisfy $(3/2)^2 + (20/3)^2 = 1681/36 = (41/6)^2$, and the area is $\frac12 \cdot \frac32 \cdot \frac{20}{3} = 5$. The number $1$ is not congruent; this is Fermat's right triangle theorem. A positive integer $n$ is congruent if and only if the elliptic curve $y^2 = x^3 - n^2 x$ has a rational point with $y \ne 0$. Equivalently, this curve has positive rank. Jerrold Tunnell gave in 1983 an explicit criterion by counting integer solutions of certain ternary quadratic equations. The criterion is necessary for $n$ to be congruent, and its sufficiency follows from the Birch and Swinnerton-Dyer conjecture. So the problem is open in general.