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I've gone with the approach of letting $n$ be either even or odd. I was able to solve for the case when $n$ is even but i don't know how to approach the case were $n$ is odd. I've done the scratch work but I don't know how to word it or how to make it work. Please help.

TRUSKI
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P.Sanchez
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3 Answers3

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Let $d=\gcd(n,n+2)$. Then $d |n$ and $d |n+2$. So $d$ divides their difference $2$.

Anurag A
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Suppose $k\geq2$ and $k$ divides $n$. Then $n\equiv0$ mod $k$, so that $n+2\equiv 2$ mod $k$. When is $2$ equivalent to $0$ mod $k$?

TomGrubb
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  • I originally did my proof this way for my Homework but my professor wanted me to prove it using two cases when n is odd and when n is even. – P.Sanchez Nov 16 '17 at 06:29
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$$\gcd(n,n+2)=gcd(n,(n+2)-n)=\gcd(n,2)$$

Now, we can consider the two cases of

  • $n$ is odd
  • $n$ is even.
Siong Thye Goh
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