Exercise 1656Number theory › Divisibility and gcdOlympiadAdded 3 Nov 2026

Primitive Pythagorean triples

#Pythagorean triples#coprime#parity

Statement

Show that in any primitive Pythagorean triple $(a,b,c)$ with $a^2+b^2=c^2$ and $\gcd(a,b,c)=1$ , exactly one of $a,b$ is even, and $c$ is odd.

Hint

Parity analysis modulo $4$ : squares mod $4$ are $0$ or $1$ .

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Both odd? If $a,b$ both odd, $a^2+b^2\equiv1+1\equiv2\pmod4$ . But $c^2\equiv0$ or $1\pmod4$ . Contradiction.

Both even? Then $\gcd(a,b)\ge2$ , contradicting primitivity.

Exactly one even. So exactly one is even; WLOG $a$ even. Then $a^2\equiv0\pmod4$ , $b^2\equiv1\pmod4$ , so $c^2\equiv1\pmod4$ , $c$ odd.

Conclusion. In any primitive triple, exactly one leg is even and the hypotenuse is odd.

Parametrization. $a=2mn$ , $b=m^2-n^2$ , $c=m^2+n^2$ with $m>n$ , coprime, opposite parity.