Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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
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 $\geq n-1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.