<s>[INST] 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?
 [/INST] $$a^2+c^2=b^2$$
$$\implies a^2+a+b=b^2$$
$$\implies \Big(a+\frac{1}{2}\Big)^2= \Big(b-\frac{1}{2}\Big)^2$$
$$\implies a+\frac{1}{2}=b-\frac{1}{2}$$
$$\implies b=a+1$$
Hence,
$$2a+1=c^2$$
Therefore, $c$ is odd, let $c=2k+1$. Putting it in above equation, you get, $$a=2k^2+2k$$
$$\implies b=a+1=2k^2+2k+1$$
This is the required general solution.
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