Find $\operatorname{lcm}(2n-1,2n+1)$ I'm trying to find the formula for $\operatorname{lcm}(2n-1,2n+1)$ with $n \in \mathbb{Z}$.
Here is my solution but I'm not sure about it.
We know that
$$\operatorname{lcm}(a,b)=\frac{\lvert ab \lvert}{\gcd(a,b)}$$
Now if we substitute we have:
$$\operatorname{lcm}(2n-1,2n+1)=\frac{\lvert (2n-1)(2n+1) \lvert}{\gcd(2n-1,2n+1)}=\frac{\lvert (4n^2-1) \lvert}{\gcd(2n-1,2n+1)}$$
Now we compute $\gcd(2n-1,2n+1)$, We can notice that
$$2n+1=(2n-1)+2$$
Thus the 2 arguments are not divisible (unless $n=0$) hence $\gcd(2n-1,2n+1)=1$
We can conclude that
$$\operatorname{lcm}(2n-1,2n+1)=\lvert (4n^2-1) \lvert$$

Yes, your work is fine. A slightly different formulation of your study of the greatest common divisor would be to notice that the gcd of two numbers divides their difference, which is $2$. Hence, the gcd is either $1$ or $2$, and the fact that they are both odd implies that it's $1$.
