1. ## Elementary # Theory

Ok,
so here s my problem, again

when proving Fermat's Theorem, we begin by considering the first p-1 positive multiples of a, that is the integers,
a, 2a, 3a, .......(p-1)a

None of there # is congruent modulo p to any other, nor is any congruent to zero, indeed, if it happend that
r*a= s*a (mod p) 1<= r < s <= p-1

then a could be canceled to give r=s (mod p), which is impossible . therefore, the previous set of integers must be congruent modulo p to 1,2,3,....p, taken in some order. Multiplying all these congruences together, we find ...

......

So my problems is how can you tell that "the previous set of integers must be congruent modulo p to 1,2,3,....p, taken in some order"

Thanks very much, guys.
Felix

2. You should know that every number is congruent to exactly one of $\displaystyle 0, 1, 2, \hdots, m -1$ modulo m.

Showing that no two integers of $\displaystyle a, 2a, 3a, \hdots, (p-1)a$ are congruent (mod p) must mean that their least residues are not congruent and therefore must all be different, i.e. some order of $\displaystyle 1, 2, \hdots, p-1$ - the least residues mod p.