Posted by Jane on Friday, November 23, 2007 at 5:26pm.
Part a:
(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:
1*2*3*4*...*(q-1) =
p*(2p)*(3p)*(4p)*...*(q-1)p.
We can write:
p*(2p)*(3p)*(4p)*...*(q-1)p =
p^(q-1) *1*2*3*...*(q-1)
And it follows that:
p^(q-1) = 1
I b)
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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