📂Number Theory

One of the Pythagorean Numbers Must Be Even

Theorem ¹

When natural numbers $a,b,c$ satisfy $a^2 + b^2 = c^2$, either $a$ or $b$ is even.

Explanation

Interestingly, one of the Pythagorean numbers must be even.

Proof

Since the square of an even number is even and the square of an odd number is odd, if $c^2$ is odd, then $a^2$ or $b^2$ must be even. If we assume $c^2$ is even, then $a^2$ and $b^2$ are both odd or both even, and it suffices to examine only the case where both are odd.

For some natural numbers $x,y,z \in \mathbb{N}$, let us define $a,b,c$ as follows. $$ a := 2x +1 \\ b : = 2y + 1 \\ c = 2z $$ Substituting these into $a^{2} + b^{2} = c^{2}$ gives $$ (2x+1)^2 + (2y+1)^2 = (2z)^2 $$ Expanding the squares gives $$ 4x^2 + 4x +1 + 4y^2 + 4y +1 = 4z^2 $$ Dividing both sides by $2$ gives $$ 2 \left( x^2 + x + y^2 + y \right) +1 = 2z^2 $$ Here the left-hand side is odd while the right-hand side is even, which is a contradiction, and therefore either $a$ or $b$ must be even.

■