A conjecture about Pythagorean triples

I noticed for the integer solutions of $a^2 + b^2 = c^2$, there don’t seem to be cases where both a and b are odd numbers. In fact, I saw this property pop up on a nice question, which required you to prove it. So I have tried proving it, but I have failed so far. A reductio ad absurdum method seems to me the most likely way of proving this.

So far, I have shown that if both a and b were odd, then $a^2 + b^2 = 4\cdot y^2$ (where y is some integer). This is because an odd squared is still odd and 2 odds make an even, and if $c^2$ is even, then it must have a factor of 4.

However this isn’t too great and my brain seems to be rather uncreative today. Any tips on how I should proceed?

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$$(2m+1)^2+(2n+1)^2 = 4(m^2+m+n^2+n)+2$$ which is even (a multiple of $2$) as you noticed but not a multiple of $4$ and so not a square.

Notice that $x^2+y^2\equiv0\pmod4$ has no integer solutions, as a square is either congruent to $0$, or to $1,$ modulo $4.$
Hope this helps.

You won’t get very far just by considering whether numbers are even or not. Note that $$(2n+1)^2=8 \cdot\frac{n(n+1)}2+1$$ and see if that helps you to take it further.

Let’s just go over the four possible squares modulo 4:

$$(4n)^2=4\cdot n^2 + 0 \equiv 0 \pmod 4$$
$$(4n+1)^2=4\cdot (4n^2+2n)+1 \equiv 1 \pmod 4$$
$$(4n+2)^2=4\cdot (4n^2+4n+1)+0 \equiv 0 \pmod 4$$
$$(4n+3)^2=4\cdot (4n^2+6n+2)+1 \equiv 1 \pmod 4$$

So here we observe that all squares are either 0 or 1 modulo 4

Note also that even squares are 0 and odd squares are 1 modulo 4.

You can have the sum of two squares essentially in three different ways in terms of parity in modulo 4: (A) the sum of two odd squares, (B) the sum of one odd square and one even square, and (C) the sum of two even squares.

(A)$\underline{\text{If you have the sum of two odd squares}}$, you are left with a sum of $\ 2 \ \pmod 4$ So you know that’s not a possibility in a Pythagorean triple.

(B) If you have one even and one odd, that adds to $ \ 1 \ \pmod 4$ so that IS a possibility. I will show that it’s essentially the only one below:

(C) The sum of two even squares will add to $ \ 0 \pmod 4$, a possible square, however, we would then be able to divide the whole equation by 4, and repeat if necessary, and the result may or may not be divisible by 4:

$$(4n)^2 + (4n+2)^2 \equiv 0 \pmod 4 \equiv 4\cdot p \pmod 4 \ \ | \ p \in \mathbb{Z}$$

$$4\cdot n^2 + 4\cdot (4n^2+4n+1) \equiv 4\cdot p \pmod 4$$

$$n^2 +(4n^2 +4n+1) \equiv \ p \pmod 4$$

Now $p$ could either be $0$, $1$, $2$, or $3$, and if it’s $2$ or $3$, then the sum won’t add to a Pythagorean Triple. And if it adds to $0 \pmod 4$, you know you can just divide by 4 again. The only thing left is if $p=1$, and if so then one of the summands must be odd, and the other must be even. This is why you observe that every pythagorean triple in its primitive state has one even leg and one odd leg.

The square of an odd number is always a multiple of four plus one: $(2n+1)^2=4(n^2+n)+1$.

So the sum of two odd squares cannot be a multiple of four.