The three sides of a right-angled triangle have lengths $a$, $b$ and $c$, where $a, b, c \in \mathbb{Z}$ - AQA - A-Level Maths Pure - Question 6 - 2019 - Paper 3

Assume that $a$ and $b$ are odd. Let:

• $a = 2m + 1$
• $b = 2n + 1$

Using Pythagoras' theorem, we find:

$c^2 = a^2 + b^2$

Substituting the values gives us:

$c^2 = (2m + 1)^2 + (2n + 1)^2$

Expanding this yields:

$c^2 = (4m^2 + 4m + 1) + (4n^2 + 4n + 1)$

$c^2 = 4m^2 + 4m + 4n^2 + 4n + 2$

Factoring out a 2:

$c^2 = 2(2m^2 + 2m + 2n^2 + 2n + 1)$

This implies that $c^2$ is even. Since the square of an odd number is odd ($c$ being odd means $c^2$ would be odd), it follows that all three cannot be odd concurrently. Therefore, it is not possible for all $a$, $b$, and $c$ to be odd.