Question about congruences and orders

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Ch1ronTL34
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The question is:

Show that if p is an odd prime and ord(p^a)a=2t, then
a^t== -1 mod p^a

First, I used ord(p^a)a to mean "order of a, mod p^a" and the == sign means congruent.

So first, I tried a few examples. Let p=3, a=2
Since ord(9)2=6, then t=3 and:
2^3 == -1 mod 9 TRUE

I continued with different values of p and a. Here is a table(sorry it looks weird):

p--a--t--p^a
3--2--3--9
5--2--10--25
11--2--55--121
13--2--78--169
17--2--68--289

It seems that p|t in all of my examples but I'm stuck...THANKS!
 
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If we take a^t and square it what do we get? Now, since a^t is not 1, since 2t is the order of a, what do you need to show?
 
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