Find $\operatorname{lcm}(2n-1,2n+1)$

Question

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 $$\operatorname{lcm}(2n-1,2n+1)=\lvert (4n^2-1) \lvert$$

Answer 1

$$\text{lcm}(2n-1,2n+1) = \frac{(2n-1)(2n+1)}{\gcd(2n-1,2n+1)}$$

Note that $\gcd(a,b) = \gcd(a-b,b)$ if $a>b$, and $\gcd(a,b) = \gcd(a,b-a)$ if $a<b$.

Since $2n+1 > 2n-1$:

$$\frac{(2n-1)(2n+1)}{\gcd(2n-1,2n+1 - (2n-1))} = \frac{(2n-1)(2n+1)}{\gcd(2n-1,2)} = (2n-1)(2n+1)$$

Answer 2

Your logic is totally correct except that the logic for the gcd can be better written as follows:

$(2n - 1)$ and $(2n + 1)$ are consecutive odd integers and hence they are co-prime i.e. $\gcd(2n - 1, 2n + 1) = 1$.

Answer 3

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 $1$.