Congruency and Divisibility of Odd in Z

[knowLittle]

Homework Statement

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

The Attempt at a Solution

I don't think I understand the question.

The few notes I have state d| (n+2 )- n

This resembles n+2 $\equiv$ n mod d , but I don't see the connection.
The congruency means that n+2 and n share the same remainder.
Any help?

 

[Dick]

Homework Statement

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

The Attempt at a Solution

I don't think I understand the question.

The few notes I have state d| (n+2 )- n

This resembles n+2 $\equiv$ n mod d , but I don't see the connection.
The congruency means that n+2 and n share the same remainder.
Any help?

The problem doesn't have anything to do with congruency. If d is a common divisor then d divides n and d divides n+2, so it follows that d must divide (n+2)-n. Don't you agree? Now use that to solve the problem.

 

[knowLittle]

Thank you. I agree.
I understand that
$d | n+2 -n$

$n+2 -n = dq,$ where q $\in Z$
Since, 2 = dq, so d and q could be 1 or 2.

 

[Dick]

Thank you. I agree.
I understand that
$d | n+2 -n$

$n+2 -n = dq,$ where q $\in Z$
Since, 2 = dq, so d and q could be 1 or 2.

Correct, that d must be 1 or 2. Now can you prove the original problem?

 

[knowLittle]

I thought that I already did.
d |n and d| n+2
n =dk and n+2 = dp, k and p are in Zn is odd, so n =2a+1 for some a in Z.

dk +2 =dp
2a+1 +2 = dp
2(a+1) + 1 = dp

I think this means that I am confused.

 

[Dick]

I thought that I already did.
d |n and d| n+2
n =dk and n+2 = dp, k and p are in Z


n is odd, so n =2a+1 for some a in Z.

dk +2 =dp
2a+1 +2 = dp
2(a+1) + 1 = dp

I think this means that I am confused.

Yeah, it does look like you have lost your way. It's much simpler than that. You know d=1 or 2. You are given that n is odd. d divides n. Can d be 2?

 

[knowLittle]

An even number can never divide an odd number. 'd' must be 1. :>

Thanks.