Show that for all positive integers n, $\text{hcf}(6n + 8, 2n + 3) = 1$

Question

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 $1.$

Note: In India and the Southeast countries, HCF is preferred to GCD... maybe that's why OP used it.. — Panglossian Oporopolist, Mar 13 '15 at 12:31
You can also note that any common factor divides $3(2n+3)-(6n+8)=1$. — Kim Jong Un, Mar 13 '15 at 12:36

score 1 · Answer 1 · answered Mar 13 '15 at 12:22

1

Yes, your proof is simple and it works.

answered Mar 13 '15 at 12:22

ghosts_in_the_code

2,022

score 0 · Answer 2 · answered Mar 13 '15 at 12:30

You can do what you've done; it's correct. Another way, but somewhat informal/intuitive...

$$6n + 8 = 2(3n+4)=6(n+1)+2=2(3(n+1)+1) => Even$$
$$2n + 3 = 2(n+1) + 1 => Odd$$

Now, it's not necessary that gcd(a,b) = 1 when a,b are odd and even respectively, but since they don't have a common factor after factorizing out, they're coprime.

Your proof is rigorous, so it's fine really.

Show that for all positive integers n, $\text{hcf}(6n + 8, 2n + 3) = 1$

2 Answers2