Determining all right-angled triangles with sides $a$, $a+k$ and $a+2k$ 
Determine all triangles that are right-angled which has the sides:
  $$a, \quad b = a+k, \quad c = a+2k.$$
  (a, k are natural numbers!)

So does this mean I have to take a look at:


*

*$a^2+b^2=c^2 \iff a^2-2ak-3k^2=0$

*$b^2+c^2=a^2 \iff a^2+6ak+5k^2=0$

*$a^2+c^2=b^2 \iff a^2+2ak+3k^2=0$ 
If so, I get (1.1) $a=3k$ (I did $f(a)=a^2-2ak-3k^2=0$)
in all other cases I get negative $k$ or negative $a$.
What's the best way to deal with this (I'll assume) simple task?

Yes! Sure, in fact
$$b^2+c^2=a^2 \iff a^2+6ak+5k^2=0$$ means $$ a = -k,  a = -5k$$
while
$$a^2+c^2=b^2 \iff a^2+2ak+3k^2=0$$ have a unique integer solution $$ a = k = 0.$$
So $$a^2+b^2=c^2 \iff a^2-2ak-3k^2=0$$ is the unique possible answer: $$ a = -k,  a = 3k.$$
Finally the sides of all the possible triangles are of the form $(3k,4k,5k)$, the simplest Pitagorean triple!
