A question about congruence in a cyclotomic coset

Question

Let $q\neq 2$ be a prime number. Let $\gcd(q^m-1,e)=1$ for all $1\leq e \leq d-1$.

Proposition: If $1\leq i<j \leq d-1$, then $j \not\equiv q^si \pmod{q^m-1}$ for any integer $s\geq 0$.

Proof: Otherwise, we would have $j-i \equiv (q^s-1)i \pmod{q^m-1}$ which results in $j-i \equiv 0 \pmod{q-1}$ which is a contradiction to the condition $\gcd(q^m-1,j-i)=1$.

Two questions:

1. How the result $j-i \equiv 0 \pmod{q-1}$ is obtained?

2. In which part of proof the condition $q\neq 2$ is used?

Thanks in advances

Source: Proposition 8.1.12 in Page 164 in Coding Theory A First Course.

Answer 1

Good questions:

1. Since $(q-1) \mid (q^m-1)$, the congruence $j-i \equiv (q^s-1)i \pmod{q^m-1}$ implies the congruence $j-i \equiv (q^s-1)i \pmod{q-1}$. However, $(q-1)\mid(q^s-1)$ as well, and so this latter congruence is equivalent to $j-i \equiv 0i \equiv 0 \pmod{q-1}$.

2. The conguence $j-i \equiv 0 \pmod{q-1}$ proves that $(q-1)\mid(j-i)$; since $(q-1)\mid(q^m-1)$ as well, we conclude that $(q-1)\mid\gcd(q^m-1,j-i)$, and therefore that $\gcd(q^m-1,j-i) \ge q-1$. However, this only contradicts $\gcd(q^m-1,j-i)=1$ when $q-1>1$, which is where the assumption $q\ne2$ is used.