Posted by Jane on Friday, November 23, 2007 at 5:26pm.
I'm not sure how to start part(a). I believe part (b) is correct, but just in case I would like to make sure it's correct.
(a) Explain p^(q-1) = 1 modq , If p and
q are distinct primes.
Ans: I was thinking of doing
something like part (b) but I
felt it was somehow wrong
because I wouldn't know how to
go from there.
(b) Determine 17^(98) mod 7 (give
answer in mod 7)
Ans: a=17, p=7 a^(p-1)=1 mod p
==> 17^6 = 1 mod 7
^6 = 
^98= [(17)^6]^(16) * (17)^2
^16 * ^2 = 289
Therefore, 17^98 = 289 mod7
***Discrete Math *** - Count Iblis, Friday, November 23, 2007 at 5:44pm
(all expressions are Mod q)
Consider the q-1 numbers p, 2p, 3p, 4p, 5p,...,(q-1)p.
All these numbers are different and nonzero. If ap = bp and p and q don't have divisors in common then a = b. So, a - b must be zero or a multiple of q, which means that all the q - 1 multiples of p are different and nonzero.
Since there are only q-1 nonzero numbers Mod q:
1, 2, 3, ... q-1,
this means that the numbers
p, 2p, 3p, 4p, 5p,...,(q-1)p
are just the numbers
1, 2, 3,... q-1
but in some different order.
This means that:
We can write:
And it follows that:
p^(q-1) = 1
You can simplify a but more by using that 17 = 3
So, you get 3^2 = 9 = 2 as the answer.
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