Is $\gcd(n, n+2) = 1$ or $2$ for any integer $n$?

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Discussion Overview

The discussion revolves around the properties of the greatest common divisor (gcd) for various expressions involving integers, specifically examining whether certain statements about gcd are true or false. The scope includes mathematical reasoning and exploration of number theory concepts.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants assert that for any integer $n$, $\gcd(n, n+1) = 1$ is true.
  • Others argue that for any integer $n$, $\gcd(n, n+2) = 2$ is false, providing the example $\gcd(3, 5) = 1$ as a counterexample.
  • Some participants propose that for any integer $n$, $\gcd(n, n+2)$ can be either $1$ or $2$, depending on whether $n$ is odd or even, respectively.
  • There is a discussion about the identity $\gcd(n, n^2+m) = \gcd(n,m)$, with some participants agreeing on its correctness and providing reasoning based on the properties of gcd.

Areas of Agreement / Disagreement

Participants generally agree on the truth of statement (a) and the falsehood of statement (b), while there is some consensus on statement (c) being true, though it is based on the parity of $n$. The discussion remains unresolved regarding the implications of these gcd properties.

Contextual Notes

Participants rely on specific examples and identities related to gcd, but there are no formal proofs provided, and the discussion does not resolve all aspects of the claims made.

Who May Find This Useful

Readers interested in number theory, properties of gcd, and mathematical reasoning may find this discussion relevant.

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Decide which of the following is true or false. If false, provide a counterexample.

(a) For any integer $n$, $\gcd(n, n+1) = 1$.

(b) For any integer $n$, $\gcd(n, n+2) = 2$.

(c) For any integer $n$, $\gcd(n, n+2) = 1$ or $2$.

(d) For all integers $n, m:$ $\gcd(n, n^2+m) = \gcd(n,m)$.

I think (a) is true and (b) is false since $\gcd(3,5) = 1.$
 
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Guest said:
I think (a) is true and (b) is false since $\gcd(3,5) = 1.$
I agree.

Guest said:
(d) For all integers $n, m:$ $\gcd(n, n^2+m) = \gcd(n,m)$.
This can be answered using the identity $\gcd(a,b)=\gcd(a,b-a)$.
 
Evgeny.Makarov said:
I agree.

This can be answered using the identity $\gcd(a,b)=\gcd(a,b-a)$.
Thank you. Is it

$\gcd(n, n^2+m) = \gcd(n, n^2+m-n) =\gcd(n, n^2+m-n-n)= \cdots = \gcd(n, n^2+m-\ell)$ where $\displaystyle \ell = \sum_{k=1}^{n}n = n^2$?
 
Yes, your reasoning for (d) is correct.
 
Evgeny.Makarov said:
Yes, your reasoning for (d) is correct.
For (c) I get $\gcd(n, n+2) = \gcd(n, 2)$ which is $2$ if $n$ is even, or $1$ if $n$ is odd. So it's true, I think.
 
Guest said:
For (c) I get $\gcd(n, n+2) = \gcd(n, 2)$ which is $2$ if $n$ is even, or $1$ if $n$ is odd. So it's true, I think.
Yes.
 

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