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

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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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