Show that for all positive integers n, $\text{hcf}(6n + 8, 2n + 3) = 1$ To show that for all positive integers $n$, $\operatorname{hcf}(6n + 8, 2n + 3) = 1$
Can I just show that since n is a positive integer, so $6n + 8 > 4n + 5$.
When we apply long division
$6n + 8 = 1(4n + 5) + (2n +3)$
$4n + 5 = 1(2n + 3) + (2n +2)$
$2n + 3 = 1(2n + 2) + 1$
Therefore the highest common factor of $6n+8$ and $4n+5$ is $1.$

Yes, your proof is simple and it works.
