Quadratic Congruence and Primitive Roots
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1. If g is a primitive root of p, show that two consecutive powers of g have consecutive least residues. That is, show that there exists k such that g^(k+1)=g^k+1(mod p) (Fibonacci primitive root)
2. Show that if p=12k+1 for somek , then (3/p)=1
3. Show that if a is aquadratic residue (mod p) and ab=1(mod p) then b is a quadratic residue (mod p)
4. Suppose that p=1+4a, where p and q are odd primes. show that (a/p)=(a/q).
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Solution Summary
This solution provides a step-by-step explanation of how to solve the given problems regarding positive roots and quadratic congruences.
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Please see the attached file for the complete solution response.
1. If g is a primitive root of p, show that two consecutive powers of g have consecutive least residues; that is, show that there exists a k such that (please see the attached file).
Proof: Note that the property does not hold for (please see the attached file) since the only primitive root of 2 is 1 and (please see the attached file).
Now suppose p is an odd prime and g is a primitive root of p. Then g is a least residue modulo p and the order of g modulo p is (please see the attached file) that is, (please see the attached file) and the powers of g, (please see the attached file) are a permutation of least residues 1, 2, ..., modulo p. In ...
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