For what value of $k$ is $(k,\ k+1,\ k+2)$ a Pythagorean triple?

Question

Aicharya · Accepted Answer

$k^{2} + (k+1)^{2} = (k+2)^{2} \Rightarrow k^{2} - 2k - 3 = 0 \Rightarrow (k-3)(k+1) = 0$. The positive root is $k = 3$, giving $(3, 4, 5)$.

Solution