Why can't a Pythagorean triple triangle have an odd area?

Question

To my understanding that a primitive triple $x$ and $y$ can be written as $x = q^2 - p^2$ while $y=2pq$ for relatively prime opposite parity $q > p$ then the area can be calculated as: $pq(q^2 - p^2) = pq(q+p)(q-p)$. Am I missing something obvious that helps prove that a Pythagorean Triple triangle cannot have an odd area?

Answer 1

As you say, we can write $x = q^2 - p^2$ and $y = 2pq$ where $p, q$ are relatively prime with different parity. Now the area of the triangle is half of $xy$. If this area is odd, then $xy \equiv 2 \; (\textrm{mod } 4)$ and \begin{align} 2pq(q^2 - p^2) \equiv 2 \; (\textrm{mod } 4) \end{align} which implies $pq(q^2 - p^2) \equiv 1 \; (\textrm{mod } 2)$. This is impossible because one of $p$ or $q$ must be even.

Let me know if I have made any errors :)

Answer 2

If $a^2+b^2=c^2$ with integers $a,b,c$, recall that the square of an even number is $\equiv 0\pmod4$ and the square of an odd number is $\equiv1\pmod8$. Therefore, either $a,b,c$ are even (and then $\frac12ab$ is even) or $c$ and one of $a,b$ are odd and the other is a multiple of 4 (because its square is a multiple of 8), so again $\frac12ab$ is even.

Answer 3

We start off with the fact that any pythagorean triples can only consist of all even numbers or $2$ odd numbers and $1$ even number.

Let the numbers are denoted $x, y, z \in \mathbb{N}$ with $z$ being the hypotenuse, and $x = p^2 - q^2$ and $y = 2pq$. Given that $y$ must be even, the only following cases arise:

If $x \equiv 0 \mod 2$ and $y \equiv 0 \mod 2$

Then, let $x = 2k, y = 2j$ for some $k, j$.

It is trivial to see $\big(\frac{1}{2}xy = 2kj\big) \equiv 0 \mod 2$.

If $x \equiv 1 \mod 2$ and $y \equiv 0 \mod 2$

Then, $x = (p + q)(p - q) \equiv 1 \mod 2 \implies \begin{cases} (p + q) \equiv 1 \mod 2 \\ (p - q) \equiv 1 \mod 2 \end{cases} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad \implies \begin{cases} p \equiv 0 \mod 2, \qquad q \equiv 1 \mod 2 \\ p \equiv 1 \mod 2, \qquad q \equiv 0 \mod 2\end{cases}$

It is inevitable for $(y = 2pq) \equiv 0 \mod 4$ for either $p$ or $q$ even.

Hence, let $y = 4n$ for some $n$.

Again, we see $\big(\frac{1}{2}xy = 2xn\big) \equiv 0 \mod 2 \quad \square.$