Congruent number elliptic curve ($E_D:D{y}^2=x^3-x$)

These elliptic curves encode the congruent number problem through rational right triangles.

For a positive square-free integer $D$, rational points on $E_D:D{y}^2=x^3-x$ with $2P\ne O$ correspond to rational right triangles of area $D$. The point $(x,y)$ gives side lengths

For an odd prime $p$, let $E_p:y^2=x^3-p^{2}x$. Its two-isogeny descent partner is $E_p^{\prime }:y^2=x^3+4p^{2}x$. The first two-torsion square-class homomorphism has image exactly $\{1,-1,p,-p\}$: prime-support bounds allow only these classes, and the four rational two-torsion points realize all of them. Every nonidentity point on $E_p^{\prime }$ has positive $x$, so its square-class image is contained in $\{1,2,p,2p\}$. The two-isogeny rank formula gives $2^{\mathrm{rank}{E}_p}=\#{\alpha }^{\prime }(E_p^{\prime })\le 4$, hence the rank is at most $2$.

For nonzero integral $D$, the elliptic curve $E_D:y^2=x^3-D^{2}x$ has rational torsion precisely $O,(0,0),(D,0),(-D,0)$. At every good prime $p\equiv 3(\mathrm{mod}4)$, cancellation of the Legendre symbols at $x$ and $-x$ gives $\#E_D({\mathbb{F}}_p)=p+1$. The torsion-freeness of the formal group over Qp for odd p makes the rational torsion subgroup inject into this group. The Dirichlet theorem on primes in arithmetic progressions excludes odd factors of its order and bounds its two-primary part by four. The four displayed rational 2-torsion points then give equality.

A positive integer is a congruent number when it is the area of a right triangle with rational side lengths. For square-free $D$, this is equivalent to the congruent number elliptic curve having a Mordell-Weil group of positive rank.

A primitive Pythagorean triple of area twice a square would satisfy $mn(m-n)(m+n)=2s^2$. Pairwise coprimality makes the odd factors squares and the even factor twice a square. The possibility $m$ even is excluded modulo eight. Thus $m=a^2,n=2b^2,m+n=c^2,m-n=d^2$. The coprime integers $u=(c+d)/2,v=(c-d)/2$ have $uv=b^2$, hence are themselves squares, and $a^2=u^2+v^2$ contradicts fourth-power sum cannot be a square. Clearing denominators gives the assertion for rational triangles.

There is no positive integer right triangle whose area is a square. For a hypothetical primitive triangle of minimal hypotenuse, its primitive Pythagorean parametrization makes $m,n,m-n,m+n$ pairwise coprime squares. Write $m=r^2$, $n=s^2$, $m+n=u^2$, $m-n=v^2$. Modulo eight forces $m$ odd and $n$ even. The new integer legs $(u+v)/2,(u-v)/2$ have hypotenuse $r$ and area $(s/2{)}^2$, a smaller positive hypotenuse, contradicting infinite descent. Clearing denominators extends the conclusion to rational right triangles of area one.

For $D>0$, a nontorsion rational point on the congruent number elliptic curve $y^2=x^3-D^{2}x$ gives infinitely many positive rational right triangles of area $D$. Apply the formulas $a=|(x^2-D^2)/y|$, $b=|2Dx/y|$, $c=|(x^2+D^2)/y|$ to its multiples. The map is finite-to-one: if $r=c/b$ is fixed, then $|x|$ satisfies $X^2-2DrX+D^2=0$, leaving only finitely many coordinate signs. Thus infinitely many points give infinitely many distinct triangles.

For a prime $p\equiv 3(\mathrm{mod}8)$, apply two-isogeny descent to the congruent number elliptic curve $y^2=x^3-p^{2}x$ and its companion $y^2=x^3+4p^{2}x$. The first square-class image is exactly $\{1,-1,p,-p\}$, already supplied by 2-torsion. The second has only positive candidates $1,2,p,2p$. The quartic covering in a two-isogeny descent equations for $p$ and $2p$ are impossible modulo $p$, because $-1$ is not a quadratic residue. For $2$ the equation is $W^2=2U^4+2p^2{V}^4$. Coprimality forces $U,V$ odd and $W=2Z$; modulo sixteen this gives $2Z^2=1+9=10$, which is impossible. Thus the second image is trivial and the square-class index formula for two-isogeny descent gives rank zero. The same prime-reduction argument bounds its rational torsion by four, and it already has four rational 2-torsion points, so it has no rational point corresponding to a right triangle of area $p$.

For $E_5:5y^2=x^3-x$,The torsion points are $O,(0,0),(1,0),(-1,0)$, and one generator of the free part is $(-4/5,6/25)$.