Prove that all multiples of 4 can appear in a Primitive Pythagorean Triple

Question

I have attempted by using the fact that the even numbers $b$ in a PPT $(a,b,c)$ can be generated using $ b = \frac{s^2 - t^2}{2} $ for any odd integers $s >t \ge 1$ that share no common factors.

We can express any odd numbers as $2k+1$ for some non-negative integers $k$, so:

$$\frac{(2n+1)^2 - (2m^2 + 1)}{2} = \frac{4(n^2 + n - m^2 - m)}{2} \\(2n+1)^2 - (2m^2 + 1)= 2(n(n+1)-m(m+1)) \\ \text{So }\frac{ s^2 - t^2}{2} = 2(2k)=4k \text{ for some integer k }$$

But I think all I did was that I proved that all even numbers $b$ in PPT must be a multiple of 4, rather than proving that all multiples of 4 can appear as b, how would I proceed from this?

Answer 1

What you have to prove is that if I give you a $b$ that is a multiple of $4$, you can express it as $\frac {s^2-t^2}2$ with $s,t$ odd and coprime. As you did, write $b=4k$, then we are looking to have $$8k=s^2-t^2=(s+t)(s-t)$$ One way is just to let $s=t+2$. Then we get $$s-t=2\\s+t=4k\\s=2k+1\\t=2k-1$$ And we know $s,t$ are odd and coprime because if they have a common factor it must divide $s-t=2$

Answer 2

$(4n^2-1)^2+(4n)^2=(4n^2+1)^2.$

All Pyth. triplets are of the form $(\,k(a^2-b^2),\,2kab,\, k(a^2+b^2)\,)$ with positive integers $k,a,b.$

All primitive triplets have $k=1$, with $a,b$ co-prime and such that $a-b$ is odd. These conditions are necessary but are also sufficient. (Along with, obviously, $a>b.)$

If $2ab=4n$ then $ab=2n,$ and we see that the above conditions on $a,b$ for a primitive triplet are met if $a=2n$ and $b=1.$

Answer 3

All non-trivial triplets have values of $B$ that are multiples of $4$ where $B=2mn$. If $m,n$ are of like parity, Euclid's formula generates trivial triples, e.g. $f(1,1)=(0,2,2)$ or $f(2,2)=(0,8,8)$ but if they are opposite parity, i.e. if $m=2x,x\in \mathbb{N}\text{ and }n=(2y-1),y\in\mathbb{N}$ or the opposite, then $4|2(2x)(2y-1)$ or $4|(2x-1)(2y)$.