# Consecutive Integers Pythagorean Triplets

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 $$(3,4,5)$$.

• There are no more. Try solving $x^2 + (x+1)^2 = (x+2)^2$. Commented Dec 8, 2020 at 4:35
• Similar but different question Commented Dec 8, 2020 at 4:41
• @player3236 There are two solutions to the equation you showed in your comment. These are $(-1,0,1)$ and $(3,4,5)$ as I showed in my answer below. Granted one is trivial. Commented Dec 10, 2020 at 2:24

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