GCD question!!!!!!!!!!!!!!!!!!!?

There's an obvious counterexample to (ii); the stated conditions allow the case
i = j
in which case it is impossible that
ib ≠ jb (mod a)
as desired.


A couple days later:
I've been taking another look at this, assuming the most likely correction.

(i) Show that for 1 ≤ i ≤ a - 1, ib ≠ 0 (mod a).

If ib=0 (mod a), then ib=am for some positive integer m.
m=ib/a, and since gcd(a,b)=1, that means that a divides i.
But i is in the range 1 ≤ i ≤ a-1, so that is impossible, and that means the assumption that
ib=0 (mod a)
must be false. This completes the proof by contradiction.

(ii) (corrected) Show that for 1 ≤ i < j ≤ a-1, ib ≠ jb (mod a)

If ib=jb (mod a), then jb - ib = am for some positive integer m.
m=(jb-ib)/a=b(j-i)/a, and since gcd(a,b)=1, that means that a divides (j-i).
But if 1 ≤ i < j ≤ a-1, then 1 ≤ j-i ≤ a-1 and that is impossible, and again our assumption, in this case
ib=jb (mod a)
must be false. (Notice that I had to eliminate the case i=j to make this argument work).

(iii) Using (ii), show that we can find k such that kb = 1 (mod a).

For each possible value of m, 1 ≤ m ≤ a-1, (ii) establishes that mb is in a different residue class modulo a. From (i), we also know that in no case is
mb=0 (mod a)
There are only a [that's the variable a] residue classes modulo a, and one is excluded, so for each integer n such that
1 ≤ n ≤ a-1
we can find an integer m, 1 ≤ m ≤ a-1, such that
mb=n (mod a)

In particular, we can find such a value, which we call k, for the case n=1.