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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.