Prove if n, i, j are integers with 1$\leq i<j\leq n$ then $\gcd (n!i+1, n!j+1) = 1$

Question

I've started by letting the $\gcd = d$. With the goal of trying to prove $d$ must be equal to 1. Since $d$ divides both $n!i + 1$ and $n!j + 1$ then there exist integers $k$ and $m$ such that $dk = n!i + 1$ and $dm = n!j + 1$. By rearranging these equations and doing some substitution, I find that $d|i - j $and $d|j$ ai. Am I off base here?

Answer 1

Now, you were not off base. Note that since $$j-i \equiv 0 \pmod {d}$$ implies that $$n!=\frac{n!}{j-i} \times (j-i)=dm $$ So we have that $$dk=n! \times i+1 \implies dk=dmi+1$$ This implies $d=1$. We are done.