Prove that $x$ and $x+1$ are coprime numbers

Question

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

$p \mid x,x+1 \Longrightarrow p \mid 1$. – njguliyev Nov 02 '13 at 17:35 — njguliyev, Nov 02 '13 at 17:35

score 37 · Accepted Answer · answered Nov 02 '13 at 17:35

37

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

answered Nov 02 '13 at 17:35

1

"If y divides x and x+1 then it divides (x+1)−x=1." Why is this true? – Kevin Nov 02 '13 at 22:14
9

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? – Nov 02 '13 at 22:19
1

Nice! A one-liner kills the question! – amWhy Mar 06 '14 at 13:51
1

Nice and concise. – drhab Jun 12 '14 at 20:18

score 17 · Answer 2 · answered Nov 02 '13 at 17:35

17

$\gcd(x,x+1)=\gcd(x,x+1-x)=\gcd(x,1)=1$.

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

answered Nov 02 '13 at 17:35

meta_warrior

1

+1 for a general method – Ramchandra Apte Nov 03 '13 at 07:14
I feel dumb, that I didn't get to it myself. Thank you! – curioushuman Feb 13 '23 at 14:43

score 3 · Answer 3 · answered Nov 03 '13 at 03:27

3

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

answered Nov 03 '13 at 03:27

Jack M

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

3 Answers3