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Given a positve integer n, we can find infinitely many positve integers $b$ such that the $n-1$ integers in the set $\{b+1,\,2b+1,\,3b+1,\,...,\,(n-1)b+1\}$ are pairwise relatively prime.

I assume that $b+1,\,2b+1,\,3b+1,\,...,\,(n-1)b+1$ are not r.p..

Let $1\le i<j\le n-1$ and $ib+1,jb+1$ are not r.p..

Let $p$ be a prime which is a factor of both $ib+1,\,jb+1$.

Question: Why should $p$ now $p\ge n-1$ ?

Proof finish: $$ \begin{align*} &p\mid (jb+1)-(ib+1)=(j-i)b\\ \Rightarrow& p\mid j-i\\ \Rightarrow& j-i<n-1\\ \Rightarrow& p<n-1, \end{align*} $$ which is a contradiction.

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The problem with your argument is that it is possible that $i=j$, and every integer divides $0$, so the last step $p \lt n-1$ is unjustified. –  Dan Brumleve Mar 7 '13 at 10:31
    
@DanBrumleve But we are trying to prove that, if $i\neq j$, then $ib+1$ and $jb+1$ are relatively prime. So how can it happen that $i=j$? –  awllower Mar 7 '13 at 10:33
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Would you mind reformulating your question. It might be me, but I am not entirely sure what you are asking. Clearly among the numbers of the form $1 + 2 i$ there are non-coprime pairs, such as $1 + 2$ and $1 + 8$. –  Andreas Caranti Mar 7 '13 at 10:33
    
Ok I just edited it. I refer to the whole set. –  Voyage Mar 7 '13 at 10:37

1 Answer 1

up vote 3 down vote accepted

Well, you can choose $b$, so take $$b=c\prod_{p \text{prime} \atop p \leq n-2}{p}$$ where $c$ is any positive integer. This gives infinitely many $b$. The rest follows from what you have done, as the prime factor cannot divide $b$, it must be $\geq n-1$.

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