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Given $\{x \mid x > 1\}$, how do I prove that any given $x$ and $x+1$ are coprime?

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$p \mid x,x+1 \Longrightarrow p \mid 1$. – njguliyev Nov 2 '13 at 17:35
up vote 33 down vote accepted

If $y$ divides $x$ and $x+1$ then it divides $(x+1)-x=1$. Conclude.

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"If y divides x and x+1 then it divides (x+1)−x=1." Why is this true? – Kevin Nov 2 '13 at 22:14
If $y$ divides $x$ and $x+1$ then there's $a$ and $b$ such that $x=ay$ and $x+1=by$ then $(x+1)-x=(b-a)y$ so... Do you understand? – user63181 Nov 2 '13 at 22:19
Yep! Thank you. – Kevin Nov 2 '13 at 22:22
Nice! A one-liner kills the question! – amWhy Mar 6 '14 at 13:51
Nice and concise. – drhab Jun 12 '14 at 20:18


Hence $x$ and $x+1$ are coprime.

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+1 for a general method – Ramchandra Apte Nov 3 '13 at 7:14

If $x$ is a multiple of $p$, then the next multiple of $p$ is $x+p$, but that's clearly larger than $x+1$.

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Or, $ $ mod $\,p\!:\ x\equiv 0\,\Rightarrow\, x+1\equiv 1\not\equiv 0\ $ (else $\,p\mid 1-0)\ \ \ $ – Bill Dubuque May 31 '14 at 16:53

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