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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.

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 $p$.

A congruent number is a positive integer which occurs as the area of a right triangle with positive rational side lengths. Scaling a triangle by a nonzero rational factor multiplies its area by a rational square, so the essential parameter is its positive square class; integer parameters can be reduced to square-free integers. For example $6$ is congruent by the triangle $(3,4,5)$, and $5$ is congruent by $(3/2,20/3,41/6)$.

The connection with elliptic curves comes from the congruent number elliptic curve in the modelFrom a rational point with $y\ne 0$, defineDirect squaring gives $a^2+b^2=c^2$, and its area isConversely, from a positive rational triangle $(a,b,c)$ of area $N$, putSubstitution using $c^2=a^2+b^2$ and $ab=2N$ gives $y^2=x^3-N^{2}x$ and $y\ne 0$. ThusThe standard rational torsion of a congruent number curve consists of $O,(0,0),(N,0),(-N,0)$. One way to see the torsion restriction is to use good primes $p\equiv 3(\mathrm{mod}4)$. The character sum cancels at $x$ and $-x$, giving $\#E_N({\mathbb{F}}_p)=p+1$. For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies good primes with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$, so the reduction of torsion points on an elliptic curve excludes $\ell$-torsion. Good primes $p\equiv 3(\mathrm{mod}8)$ similarly bound the two-primary torsion by four, and the four displayed points give equality. Consequently the Mordell-Weil theorem turns the criterion into positive rank of $E_N(\mathbb{Q})$. Arithmetic two-descent on an elliptic curve and local obstructions are therefore tools for the congruent number problem, and a nontorsion point gives infinitely many rational right triangles from a nontorsion point by taking multiples.

We now give elementary proofs for $1$ and $2$, including the descent steps. Recall the primitive Pythagorean parametrizationwhere $m>n>0$ are coprime and of opposite parity. To derive it, in a primitive Pythagorean triple put the even leg at $A$. The relatively prime integers $(C+B)/2,(C-B)/2$ have product $(A/2{)}^2$, so each is a square, say $m^2,n^2$. This gives the formulas and the parity condition. The area is $mn(m-n)(m+n)$, and its four factors are pairwise coprime: their common divisors reduce to divisors of $m,n$ or of two, while $m\pm n$ are odd.

Area one is impossible. Clearing denominators of a hypothetical rational triangle of area one gives an integer triangle with square area. Dividing out its common factor leaves a primitive integer triangle whose area is still a rational square; an integer which is a rational square is an integer square. Choose such a triangle with smallest hypotenuse. Pairwise coprimality in the area product forcesThe positive integers $u,v$ are odd and coprime. If $m$ were even, then $n$ would be odd and $u^2-v^2=2s^2$ would be $2$ modulo eight, whereas two odd squares have difference zero modulo eight. Hence $m$ is odd, $n$ is even and $s$ is even.

Now set $a=(u+v)/2$ and $b=(u-v)/2$. These are positive coprime integers andThus $(a,b,r)$ is another integer right triangle with square area and hypotenuse $r$, strictly smaller than $m^2+n^2$. This contradicts the minimal choice and proves the Fermat right triangle theorem. In particular $1$ is not a congruent number.

For area two we first establish the auxiliary descent fourth-power sum cannot be a square. Suppose $x^4+y^4=z^2$ with positive integers, and choose a primitive solution with smallest $z$. Two odd fourth powers sum to $2$ modulo sixteen, so after exchange $x$ is odd and $y$ is even. Parametrize the primitive triangle $(x^2,y^2,z)$:The case $m$ even, $n$ odd makes $x^2\equiv 3(\mathrm{mod}4)$, so $m$ is odd and $n$ even. Coprimality then gives $m=u^2$ and $n=2v^2$. ConsequentlyThis second triangle is primitive, since $x$ is odd and $\gcd (x,v)=1$. Parametrize it as $x=r^2-s^2$, $2v^2=2rs$, $u^2=r^2+s^2$, with $r,s$ coprime. The equality $v^2=rs$ makes $r=a^2$, $s=b^2$. Thusa new positive solution with smaller hypotenuse $u<z=u^4+4v^4$. This contradicts infinite descent, proving the auxiliary assertion.

Area two is also impossible. Clearing denominators and removing a common factor now gives a primitive integer triangle with area twice an integer square. Indeed, if an integer is $2(a/b{)}^2$ with $\gcd (a,b)=1$, then $b^2$ divides two, forcing $b=1$. Its area product would satisfyThe odd factors must be squares and the unique even factor twice a square. As in the preceding modulo-eight argument, $m$ even would force $m+n=c^2$, $m-n=d^2$ and $n=b^2$ odd, with $c^2-d^2=2b^2$, impossible modulo eight. ThusThe integers $u=(c+d)/2$ and $v=(c-d)/2$ are positive and coprime, and $uv=(c^2-d^2)/4=b^2$. Hence $u=r^2$, $v=s^2$. Butcontradicting the auxiliary descent. This proves two is not a congruent number, and concludesThe two obstructions are global arithmetic descents. The elliptic-curve correspondence organizes the wider problem, but by itself is not a substitute for proving these particular nonexistence assertions.

Let $P_0=(-4,6)$, the infinite-order point established in part (c). Every positive multiple $n{P}_0$ is rational and is neither $O$ nor 2-torsion: either possibility would force some nonzero multiple of $P_0$ to vanish. Write its coordinates as $(x,y)$, so $y\ne 0$ and $x\notin \{0,5,-5\}$.

The stated algebraic identity gives positive rational sidesThey satisfy $a^2+b^2=c^2$, and their area isby the curve equation. This is the infinitely many rational right triangles from a nontorsion point construction for the congruent number elliptic curve.

The infinitely many points cannot produce only finitely many triangles. For a fixed ratio $r=c/b$, the positive number $X=|x|$ satisfies $X^2-10rX+25=0$, so there are at most two choices for $|x|$, then at most two signs of $x$ and two signs of $y$. Thus the map to ordered triangles is finite-to-one. Permuting the two legs introduces only another finite ambiguity. Consequently there are infinitely many distinct rational right triangles of area five. For $P_0$ itself the sides are $(3/2,20/3,41/6)$.

Use the congruent number elliptic curve in the equivalent coordinatesThe rational point $P=(25,100)$ lies on it, since $25^3-225\cdot 25=10000$. By the preceding rational torsion of a congruent number curve result, every rational torsion point of an elliptic curve has $y=0$ or is $O$. Thus $P$ is not a torsion point of an elliptic curve, and its positive multiples give infinitely many distinct rational points with nonzero $y$.

For any such point $(x,y)$, putThe identity $(x^2-225{)}^2+(30x{)}^2=(x^2+225{)}^2$ proves that these positive rational numbers are the sides of a right triangle. Their area isAll three sides are nonzero because a point with $y\ne 0$ has $x\notin \{0,15,-15\}$. For $P$ the construction gives $(a,b,c)=(4,15/2,17/2)$.

It remains to ensure that infinitely many points do not describe only finitely many triangles. Given the ordered positive pair $b,c$, set $r=c/b$. Then $X=|x|$ satisfiesThere are at most two possible $X$, then at most two signs of $x$ and two signs of $y$. Thus each ordered triangle has at most eight preimages; allowing interchange of its legs still gives a finite number. There are infinitely many distinct rational right triangles of area $15$. This is the infinitely many rational right triangles from a nontorsion point principle.

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:

Let a rational right triangle have legs $A,B$, hypotenuse $C$, and area $AB/2=D$. After interchanging the legs if necessary, setUsing $C^2=A^2+B^2$ and $AB=2D$ givesso $P=(x,y)$ lies on the congruent number elliptic curve $E_D:D{y}^2=x^3-x$. The triangle is nondegenerate, so $y\ne 0$ and $2P\ne O$.

Conversely the three displayed lengths in the question satisfyand their area isThus every such triangle is ${\Delta }_P$.