Consecutive Integers Pythagorean Triplets

Question

I have a question for which I was not able to find an answer online. I was wondering how many Pythagorean Triplets we have found till now which consists of three consecutive integers like $(3,4,5)$.

Answer 1

$$x^2 +(x+1)^2=(x+2)^2\\\implies x^2 +(x+1)^2-(x+2)^2=0\\\implies x^2 - 2 x - 3 = 0$$

Using the quadratic equation, we find $$x\in\{3,-1\}\implies \text{the set of triples}\quad S=\{(3,4,5),(-1,0,1)\}$$ There are no other solutions.

Answer 2

For slightly simpler numbers, let $x$ be the middle number then you have to solve $$ \begin{split} (x-1)^2 + x^2 &= (x+1)^2 \\ x^2 &= (x+1)^2 - (x-1)^2 = 2x\cdot2 \end{split} $$ which yields $x=0$ and $x=4$. Can you finish?