Question

If $n\in \mathrm{\mathbb{Z}}$,what are the possible values of

(a)$\left(n,n+2\right)$ (b)$\left(n,n+6\right)$

Short answer

(a) The possible values of $d$are 1 and 2.

(b) The possible values of $d$are 1, 2, 3, and 6.

Step-by-step solution

Definition

According to the definition, if $a,b\ne 0$, then their greatest common divisor exists with a unique value. Thus, the greatest common divisor satisfies the condition $\left(a,b\right)\ge 1$ , since 1 is the divisor of both $a,b$.

Solve for n,n+2

It is given that $\left(n,n+6\right)$.

Let $d=\left(n,n+6\right)$, then $d|2$and $d|n+2$, then there exist some integers $j$ and $k$ such that $jd=n$ and $kd=n+2$. Both the conditions imply that:

$\begin{array}{rcl}2& =& kd-jd\\ \left(n+2\right)-n& =& d\left(k-j\right)\end{array}$

Hence, $d$ is a positive integer that divides 2; the values of $d$ can be 1 and 2, as both are factors of 2.

Solve for  n,n+6

It is given that $\left(n,n+6\right)$.

Let role="math" localid="1646130808945" $d=\left(n,n+6\right)$. By following the similar procedure as in step 2, $d|6$, then the factors of 6 are 1, 2, 3, and 6 itself, then $d\in \left\{1,2,3,6\right\}$.

Hence, the possible values of are 1, 2, 3, and 6.