If none of numbers: $a,a+d,a+2d,...,a+(n-1)d$ are divisible by $n$,then prove that $n,d$ are coprime.
Since none of the given numbers are divisible by $n$,then their remainders mod $n$ are $1,2,...,n-1$.Based on pigeon hole principle I deduce that there are two numbers among them such that:
$$a+(i-1)d\equiv a+(j-1)d\pmod n,(0<i,j<n)\Rightarrow$$ $$(i-j)d\equiv 0\pmod n$$
Which means $n|d$ because $i-j<n$.What's wrong with my solution which contradicts the problem??!!