Some time ago, I posted a question on six primes numbers forming an arithmetic progression, and someone commented the following: " Let m and n be positive integers with gcd(m,n)=1. Then the sequence m,2m,3m,…,nm when taken mod n will give the integers from 1 to n , each occurring exactly once ". I wanted to know why so I tried to "prove" it, this what I did:
I supposed the contrary, that in the sequence a_1 m \equiv a_2 m \pmod n, with the condition that a_1\not \equiv a_2 \pmod n (I do not know if it´s possible to use that notation but it just means that they are not congruent) , because a_1 \neq a_2 and because both a_1 and a_2 and have to be less or equal to n .
So I did this:
a_1 m \equiv a_2 m \pmod n
a_1 m - a_2 m \equiv 0 \pmod n
(a_1 - a_2)m\equiv 0 \pmod n
This means that (a_1 - a_2)m is a multiple of n, but this is a contradiction because \gcd(m,n)=1 and since a_1\not \equiv a_2 \pmod n then a_1 - a_2 \not \equiv 0 \pmod n, which means that a_1 - a_2 doesn´t have all factors of n and therefore it is not a multiple.
This may be in a textbook, so sorry if I made you lose your time.
Anyway, I would just like you to tell me if it´s ok what I did.
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