Prove that 1 is the only "common" divisor of the integers n and n+2

Question

Let n be any odd integer. Prove that 1 is the only "common" divisor of the integers n and n+2.

I think you have to find gcd(n, n+2) and say that since n odd then then n+2 will also be odd. Thus n + 2 is either an odd prime or odd non-prime. If n is a odd non prime then, n+2 must be prime. Don't know if my reasoning is right but thats the way I am thinking about this. Any help?

What you need to notice is that any number that divides two numbers divides their difference. — Gerry Myerson, Jul 19 '14 at 12:59
Hint $,\ \gcd(n,n!+!a) = \gcd(n,a),$ by Euclid's algorithm. Put $, a = 2.\ \ $ — Bill Dubuque, Jul 19 '14 at 13:52

score 1 · Answer 1 · answered Jul 19 '14 at 13:03

1

$n$ and $n+2$ can both be composite. For instance 25 and 27 are both composite.

Let $d$ be a common divisor: $d\mid n$ and $d\mid n+2$. Then $d \mid (n+2) - n = 2$ so $d = 1$ or $d = 2$, but $n$ is odd, so $2 \nmid n$, thus, $d = 1$.

answered Jul 19 '14 at 13:03

Darth Geek

12,296

score 0 · Answer 2 · answered Jul 19 '14 at 13:26

Let d take the values of all the factors of n one by one. This implies that n/d = m, where m is some integer.

Therefore, (n+2)/d = m + (2/d)

This means that for d to be a factor of n+2, d has to be a factor of 2. But d can be only 1 or 2. And since n is an odd number, d cannot be 2.

Hence d = 1

Hope that helps :)

Prove that 1 is the only "common" divisor of the integers n and n+2

2 Answers2