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?
 A: 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: $a^2 + c^2 = b^2$
$c^2 = b^2 - a^2 = (b-a)(a+b)$ 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 $c \equiv 0 \pmod 5$.
A: $$a^2+c^2=b^2$$
$$\implies a^2+a+b=b^2$$
$$\implies \Big(a+\frac{1}{2}\Big)^2= \Big(b-\frac{1}{2}\Big)^2$$
$$\implies a+\frac{1}{2}=b-\frac{1}{2}$$
$$\implies b=a+1$$
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.
A: Substituting c^2 into the second equation we obtain a^2+a+b=b^2 to that we have a(a+1)=b(b-1), now this is a quadratic in terms of a (or b) which can be solved by the quadratic formula. So we have a^2+a+(b-b^2)=0 and solving gives a=-b or a=b-1. Now if a=-b then c^2=0 so c=0, hence we have found the trivial solution (a,b,c)=(0,0,0). 
Now if a=b-1, then from c^2=a+b=b-1+b=2b-1 we obtain that c=+-sqrt(2b-1), but since we are considering natural numbers, we can take the positive root c=sqrt(2b-1). Finally either we have the trivial solution or all solutions are of the form (a,b,c)= (b-1,b, sqrt(2b-1)) whenever sqrt(2b-1) gives a natural number, i.e 2b-1=n^2, odd square numbers. 
I believe your set of solutions are fine, just the trivial solution is missing as far as I can tell.
