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A rooted ternary tree that generates every primitive Pythagorean triple from $(3,4,5)$ by three fixed linear transformations. Keeping a consistent leg convention gives each primitive triple once; multiplying its sides then supplies all nonprimitive triples.

The formulas $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$ give a Pythagorean triple for integers $m>n>0$, because $(m^2-n^2{)}^2+(2mn{)}^2=(m^2+n^2{)}^2$. They give a primitive Pythagorean triple exactly when $m,n$ are coprime and have opposite integer parity. Every primitive Pythagorean triple arises this way after possibly exchanging the legs; scaling gives the nonprimitive triples.

Choose a primitive counterexample with minimal $z$ and $y$ even. Its Pythagorean triple $(x^2,y^2,z)$ has $x^2=m^2-n^2$, $y^2=2mn$. Modulo four forces $m$ odd and $n$ even, so $m=u^2$ and $n=2v^2$. The primitive triangle $(x,2v^2,u^2)$ then has a parametrization with $v^2=rs$ and $u^2=r^2+s^2$. Coprimality makes $r=a^2,s=b^2$, giving $a^4+b^4=u^2$ with $u<z$. This contradicts infinite descent.

There is a necessary ordering qualification: the printed formula requires the even leg to be called $b$. The primitive Pythagorean triple $(a,b,c)=(4,3,5)$ satisfies the printed hypotheses but cannot have $b=2nm$. Thus the intended assertion is true after interchanging the two legs if necessary. We prove that version, including the parity condition on the parameters.

A primitive Pythagorean triple has pairwise coprime sides: a prime dividing any two also divides the third. The two legs cannot both be odd, since their squares would sum to $2(\mathrm{mod}4)$, and cannot both be even. Call the odd leg $a$ and the even leg $b$; then $c$ is odd. The positive integerssatisfy $uv=(b/2{)}^2$ and are coprime, since a common divisor divides both $c$ and $a$. A product of coprime positive integers is a square only when both factors are squares, by their prime factorizations. Write $u=n^2$, $v=m^2$. Then $n>m>0$, $\gcd (n,m)=1$, andFurthermore $n,m$ have opposite parity, since $c$ is odd. Conversely these coprimality and parity conditions give a primitive triple: the square identity follows by expansion, and a common odd prime divisor of its legs would divide both $n,m$, while the odd leg excludes a common factor two. This is the primitive Pythagorean parametrization.

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.

For a Pythagorean triple, each square number is $0$ or $1$ modulo $3$. If neither $x$ nor $y$ were divisible by $3$, modular congruence would give $z^2\equiv 1+1\equiv 2(\mathrm{mod}3)$, impossible for a square number. Thus at least one of $x,y$ is divisible by $3$; both can be divisible, as in $(9,12,15)$.

An odd number has square number congruent to $1$ modulo $4$, whereas an even number has square number congruent to $0$. If both $x$ and $y$ were odd numbers, their sum of square numbers would again be $2$ modulo $4$, which no square number can be. Hence $x$ and $y$ cannot both be odd. They can both be even; the conclusion does not require the Pythagorean triple to be primitive.

Every primitive Pythagorean triple, after exchanging its legs, has this form with $m>n>0$, $\gcd (m,n)=1$ and opposite parity. Its area is $mn(m-n)(m+n)$. The factors $m,n,m-n,m+n$ are pairwise coprime, because $m\pm n$ are odd. Conversely these conditions give a primitive triple. The parametrization follows by factoring the relatively prime integers $(c+b)/2$ and $(c-b)/2$ when $a$ is the even leg.

A Pythagorean triple $(a,b,c)$ whose coordinates have greatest common divisor 1. Its two legs have opposite integer parity, and every Pythagorean triple is a unique positive integer multiple of a primitive Pythagorean triple up to the ordering of its legs.

The equation $a^2+b^2=c^2$. Its positive integer solutions are Pythagorean triples; primitive Pythagorean triples are those with no common factor greater than 1.

For a right triangle in Euclidean geometry with perpendicular side lengths $a,b>0$ and hypotenuse $c$, the Pythagorean theorem statesGeometrically, the area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides. Euclid's Elements, Book I, Proposition 47 states this area equality; Proposition 48 states its converse: a triangle satisfying the equality has a right angle opposite side $c$.^[ref][ref]

For $a=3$ and $b=4$, we compute $c=\sqrt{9+16}=5$. The triangle has area $ab/2=6$, and its side lengths form a Pythagorean triple. Conversely, a triangle measured to have sides $3,4,5$ must have a right angle between the sides $3$ and $4$. This makes the equality both a calculation tool and a test for perpendicularity.

The three squares in the next figure have areas $3^2=9$, $4^2=16$ and $5^2=25$ in squared length units. These are different from the central right triangle's area $3\cdot 4/2=6$. The Pythagorean theorem adds the areas of the attached squares, not their side lengths and not the area of the triangle. The square on the hypotenuse is tilted, but a rotation preserves its area. This makes the formula visible without suggesting that the hypotenuse has to align with an axis.

Historically, Euclid's Elements records an organized proof of the Pythagorean theorem as an equality of areas, rather than the modern equation between numbers representing lengths. Its placement in Book I makes it available before the treatment of similar triangles in Book VI. This is evidence of how the result was established within that text; the name Pythagorean theorem alone does not establish who first discovered or proved it. The precise historical anchor here is the surviving proposition, rather than an unsupported story of a single inventor.^[ref]

The dissection below makes the general identity tangible. Both outer squares have side $a+b$ and contain the same four congruent right triangles. In the first arrangement, their hypotenuses bound a square of side $c$: adjacent acute angles together make a right angle. In the second, the uncovered regions are squares of sides $a$ and $b$. Subtracting the same total triangle area from equal outer areas givesFor the pictured dimensions, the shared uncovered area is $49-24=25=9+16$.

The same mechanism survives in inner product spaces: for orthogonal vectors $u,v$, expanding the inner product gives $\Vert u+v{\Vert }^2=\Vert u{\Vert }^2+\Vert v{\Vert }^2$. This is why orthogonal projection separates a squared distance into independent contributions. The familiar right triangle is the simplest picture of a much more reusable principle.

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.