I am reading a solution to a problem where they have $m,n$ positive integers and there is a claim that $(n-1) \mid (n+1) \implies (n-1) \mid 2$, but I don’t know how to prove this. If we assume $(n-1) \mid (n+1)$, then $n+1=(n-1)k = nk-k$ for some integer, but how can I get the $2$ to appear here?
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6$(n+1)=(n-1)+2$. – geetha290krm Jun 15 '22 at 09:28
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$n+1=(n-1)k \implies k=1+\frac{2}{n-1}$ and you want this to be an integer – Henry Jun 15 '22 at 09:35
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3$n+1=(n-1)k$ now subtract $n-1$ from both sides; $2=(n-1)(k-1)$ – Aditya_math Jun 15 '22 at 09:35
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1It will be useful if you know the general result:if $d\mid a$ and $d\mid b$ then $d\mid ax+by$ for any integer x and y.Here take d=n-1,a=n+1,b=n-1 ,x=1 and y=-1 – Amit Kumar Basistha Jun 15 '22 at 10:11
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By here in the dupe $,n-\color{#c00}1\mid n+1\iff n-1\mid\color{#c00}1+1\ \ $ – Bill Dubuque Jun 15 '22 at 12:53
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Rename: $a=n-1$. Then consider $(n-1)\mid(n+1) \Rightarrow a\mid (a+2)$ – Keith Backman Jun 15 '22 at 17:27
1 Answers
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Observe that $(n-1)|(n-1)$ and we are given that $(n-1)|(n+1)$.
So, $(n-1)$ divides any linear combination of both and as mentioned in the comments
above,
$(n+1)-(n-1)=2$.
So, $(n-1)|2$.
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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here – Bill Dubuque Jun 15 '22 at 12:53