Let m be divisible by $1,2, … , n$.
Show that the numbers $1+m(1+i)$ where $i = 0,1,2, … , n$ are pairwise relatively prime.
My proof was as following let us have two different numbers $1+m(1+i)$ and $1+m(1+j)$, let d divides them. Thus $d\mid i-j$.
I feel this won't lead anywhere, any hints or solutions will be appreaciated.
Best Answer
Actually, what you did does lead to somewhere, although you missed a factor of $m$. The appropriate assumption is for some $i \neq j$, there's a $d \ge 2$ where $d \mid 1 + m(1 + i)$ and $d \mid 1 + m(1 + j)$, leading to $d \mid m(i - j)$. Note each prime factor $p$ of $d$ must divide $m$ and/or $i - j$. Since $|i - j| \le n$, if $p \mid i - j$ then $p \le n$, so because $2$ through $n$ divides $m$, you also have $p \mid m$. As such, in any case, all prime factors $p$ of $d$ must have $p \mid m$.
This means $p \mid m(1 + i)$, so $p \not\mid 1 + m(1 + i)$, and likewise $p \not\mid 1 + m(1 + j)$. Since this means $d$ doesn't divide either value, this is a contradiction of the assumption, thus showing no such $d \ge 2$ exists, i.e., all of the numbers are relatively prime.