If none of numbers: $a,a+d,a+2d,...,a+(n-1)d$ are divisible by $n$,then prove that $n,d$ are coprime.

Question

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??!!

Answer 1

The problem statement is incorrect. The conclusion should instead be that $n$ and $d$ are not coprime.

Your argument is not quite right, though. You know $(i-j)d$ is divisible by $n$, but this does not mean $n$ divides $d$ since $n$ may not be prime.

Answer 2

A quick counter-example $1, 1+1\cdot2, 1+2\cdot2, 1+3\cdot2=1+(4-1)\cdot2$ none divisible by $4$, but $\gcd(4,2)=2$.

Getting to the conclusion that $(i-j)d\equiv 0\pmod n$ (which is $(i-j)d=n\cdot Q_1$) is good, but not enough as it was pointed already. Let's assume that $\gcd(d,n)=1$, this means that $d$ and $n$ has no common factors. So every $p$-prime factor of $n$ will divide $i-j$ (otherwise $\gcd(d,n)\geq p >1$), concluding that $n \mid (i-j) < n$ which is a contradiction for $\gcd(d,n)=1$. So $\gcd(d,n)\ne1$.