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

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