# Finding $\gcd(2n-1,2n+1)$ . . . I think it's 1 but is my logic right?

Ok i feel like this is an easy question that I'm making more complicated than it needs to be

\begin{align} 2n+1 &= (2n-1) + 2\\ (2n-1) &= 2(n-1)+1 \end{align}

so is it $1$ ?

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@AndréNicolas: What is written might not be a proof, but if one wrote "By the Euclidean algorithm," put the calculation above, and ended with "Therefore, the GCD is $1$", I think that would suffice. Although the post has been edited since you commented, so maybe it wouldn't have before the edit. –  Aaron Jan 31 '13 at 3:08
The comment was made when there were incorrect formulas in the post. –  André Nicolas Jan 31 '13 at 3:10

I always do the same thing. Call $g = \gcd(2n-1, 2n+1).$ So $g$ divides both of htem, therefore it divides their difference, $g \mid ((2n+1)-(2n-1))$ or $g\mid2.$ Finally $g\mid 2$ and $g \mid (2n+1),$ so $g\mid (2n+1) - n \cdot 2,$ and $g\mid1$ and $g=1$

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It's simpler done equationally vs. relationally - see my answer. –  Math Gems Mar 28 '13 at 2:12

What if $\gcd(2n-1,2n+1)=d\ge2$? Then $d$ is divisible by some prime number. A prime number other than $2$ cannot divide both of two numbers that differ by $2$.

But $2$ also cannot divide both numbers since it divides neither of them.

The conclusion follows.

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$$d=(2n-1,2n+1)\Longrightarrow \exists \,x,y,\in\Bbb N\,\,\,s.t.$$

$$2n-1=xd\;\;,\;\;2n+1=yd\Longrightarrow 2=(y-x)d\ldots$$

Can you take it from here?

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$\rm mod\ (\color{#C00}{2n\!-\!1},\color{#0A0}{2n\!+\!1})\!:\,\ \color{#C00}1\equiv \color{#C00}{2n}\equiv (\color{#C00}1\!+\!\color{#0A0}1)n\equiv(\color{#C00}{2n}\color{#0A0}{\!-\!2n})n\equiv 0.\ \$ QED

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