solutions of $a+b=c^2 , a^2+c^2=b^2$ ; $a,b,c$ are natural numbers

$$a^2+c^2=b^2$$ $$\implies a^2+a+b=b^2$$ $$\implies \Big(a+\frac{1}{2}\Big)^2= \Big(b-\frac{1}{2}\Big)^2$$ $$\implies a+\frac{1}{2}=b-\frac{1}{2}$$ $$\implies b=a+1$$ 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.

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 $m+n>0,$

$$1=k(m-n)^2$$

$\implies k=1,m-n=\pm1$

$a^2 + c^2 = b^2$

$c^2 = b^2 - a^2 = (b-a)(a+b)$ but $a+b = c^2$ so if we assume $a+b \ne 0$, we have $b-a = 1$ and $b = a+1$ and we have

$a^2 + c^2 = (a+1)^2$ and $2a + 1 = c^2$

If we replace $c^2$ with $2a+1$ we have $a^2 + 2a + 1 = (a+1)^2$ which is always true. So $2a+1=c^2$ can be any number that is both; an odd number at least equal to $3$ ($a \ge 1$); and a perfect square, and $c=\sqrt {2a+1}$ is a square root of an odd perfect square greater than $3$ which can be any odd integer greater than $1$.

So for any $k \in \mathbb N$ we can have $c = 2k +1$, $a= \frac {c^2 -1}2= 2k(k+1)$ and $b = \frac {c^2 +1}2 = 2k^2 + 2k + 1$.

Those are all of them.

If $k\equiv 0,4 \pmod 5$ then $a\equiv 0 \pmod 5$. If $k\equiv 1,3\pmod 5$ then $b\equiv 0 \pmod 5$ and if $k \equiv 2\pmod 5$ then $c \equiv 0 \pmod 5$.