The problem involves proving that 1 is the only common divisor of an odd integer n and n+2. The discussion centers around the properties of divisibility and congruence in relation to odd integers.
Participants are actively engaging with the problem, with some providing insights into the nature of divisors and the implications of n being odd. There is a recognition that d must be either 1 or 2, but confusion remains regarding how to proceed with the proof.
There is an emphasis on the properties of odd integers and the implications for divisibility, with participants questioning whether an even divisor can apply to an odd integer.
knowLittle said: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?
knowLittle said: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.
knowLittle said: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.