Science:Math Exam Resources/Courses/MATH312/December 2005/Question 09
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Question 09 

Prove In this section, prove the statement given to you. Suppose p is a prime with , and r is a primitive root modulo p. Show that r is also a primitive root modulo p. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Suppose that s is the smallest power of such that . What can we say about s? 
Hint 2 

Using hint 1, we have that . As r is a primitive root, we have that and that . As these are the only possible values for a power of r to be equivalent to a power of 1, we must have that either or . Assume the second case and find a contradiction. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. Assume towards a contradiction that is not a primitive root. Then for some s smaller than . Now, we have that (multiply both sides of the previous congruence by . As r is a primitive root, we have that and that . As these are the only possible values for a power of r to be equivalent to a power of 1, we must have that either or . Suppose towards a contradiction that . Then we have that by the above that
valid since is even. This is a contradiction and hence making a primitive root. 