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

[Guest2]

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 $\gcd(3,5) = 1.$

 

[Evgeny.Makarov]

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

I agree.

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

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

 

[Guest2]

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 $\displaystyle \ell = \sum_{k=1}^{n}n = n^2$?

 

[Evgeny.Makarov]

Yes, your reasoning for (d) is correct.

 

[Guest2]

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.

 

[Evgeny.Makarov]

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.