Hi, I just registered for this site, and it looks like a pretty neat place.
I just needed a confirmation that I did a couple proofs right. I'm taking Number Theory at my university, and on Tuesday I have to teach the class two proofs.
My first one is to prove that
)
has exactly two solutions: 1 and p-1
proof: let
be any solution. SO we have
)
so
(r-1))
thus
 or p \mid (r-1))
Because p is a prime (this is important) . Correct.
otherwise expressed as
 \equiv 0(modp) or (r-1) \equiv 0(modp))
. So
 or r \equiv 1(modp))
. Since r is a least residue (modp) we have r = p -1 or 1
Correct. Note that 
...
The other one I have to do is prove that if (a,m)
doesnt divide b, then
)
has no solutions.
Proof: We can prove this by looking at the contrapositive, which is logically equivalent. So if
 \mid b)
then
has a solution.
This is not the contrapositive! It is "if 
has a solution then \mid b)
.
Fix this.
Tonio
Let r be a solution, so
)
. Thus,
)
and there exists an integer k such that
)
. Since
 \mid a)
and
 \mid (km))
it follows that
Thanks guys I appreciate it
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Originally Posted by skyzo 
