solutions of $a+b=c^2 , a^2+c^2=b^2$ ; $a,b,c$ are natural numbers So it all started with a fun observation, $12+13=5^2$ and these are Pythagorean triplets($5,12,13$), so I thought are there more such numbers? 
with brute force I was able to get $(24,25,7)$ and $(40,41,9)$.
Then I was able to find 3 families of solutions.
$(50k^2+50k+12 , 50k^2+50k+13 , 10k+5)$
$(10k+4 , 10k+5 , \sqrt{20k+9})$
$(10k, 10k+1 , \sqrt{20k+1})$
ps: I found these by using the property of Pythagorean triplets that they have at least one multiple of 5 in it.
My question is are there more sets of solution and how do I know I haven't missed any?

WLOG $a=k(m^2-n^2), b=k(m^2+n^2),c=2mnk$
We need $$2km^2=(2mnk)^2$$
$$1=2kn^2$$ which is untenable
If $a=2mnk, c=k(m^2-n^2),b=?$
$\implies k(m+n)^2=k^2(m^2-n^2)^2$
As $m+n>0,$ 
$$1=k(m-n)^2$$
$\implies k=1,m-n=\pm1$
