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 Mechanics - Question 6 - 2019 - Paper 3

Assume that $a$ and $b$ are both odd. Therefore, we can express them as:

• $a = 2m + 1$
• $b = 2n + 1$
  for integers $m$ and $n$.

Now, using the Pythagorean theorem, we have:
$c^2 = a^2 + b^2$
Substituting our expressions gives:
$c^2 = (2m + 1)^2 + (2n + 1)^2$
Expanding the squares, we get:
$c^2 = (4m^2 + 4m + 1) + (4n^2 + 4n + 1) = 4m^2 + 4n^2 + 4m + 4n + 2$
This can be factored as:
$c^2 = 4(m^2 + n^2 + m + n) + 2$
This shows that $c^2$ is even because it can be expressed as 4 times an integer plus 2, thus $c^2$ is not odd. Since $c^2$ is even, $c$ itself must also be even.
Therefore, it is impossible for all three sides $a, b$, and $c$ to be odd.