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This is the Exam of Number Theory which includes Concerning Congruences, Statement, Solutions, Infinitely Many Primes, Smallest Positive Number, Every Integer, Explain, Solve etc. Key important points are: Integer Solution, Positive Integers, System, Divides, Odd Prime Number, Infinitely Many Primes, Nonnegative Integers, Quadratic Non Residue, Algebraic Integers, Root
Typology: Exams
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Instructions: There are nine problems below (turn the sheet over for two of the problems).
Answer as many as you can on the blank pages provided with this test. You may keep the
questions when you are through.
of congruences x ≡ b 1 (mod m 1 ) x ≡ b 2 (mod m 2 )
has an integer solution if and only if b 1 ≡ b 2 (mod d).
(i) Let q be a prime number such that q divides ap^ −1. Prove that the order of a modulo q is either 1 or p.
(ii) Explain why either q|(a − 1) or q ≡ 1 (mod 2p).
(iii) Prove that there are infinitely many primes which are 1 modulo 2p.
that h!k! + (−1) h ≡ 0 (mod p).
(i) Prove that 3 is a primitive root modulo p.
(ii) Let a be a quadratic non-residue modulo p. Prove a is a primitive root modulo p.
gcd( a − 1 , 2 b − 1) = 2 d − 1.
−7). Explain why R is Euclidean.
(a) Find a polynomial g(x) that has 3/(α − 1) as a root.
(b) Is { 1 , α, α^2 } an integral basis for the ring of algebraic integers in Q(α)? Justify your answer. (Hint: You should be able to see quickly that ∆(1, α, α 2 ) is not squarefree. This means that its value cannot be used in the obvious way to answer this question, regardless of the answer. I suggest instead thinking about what part (a) has to do with the question.)
such that a 2 ≡ −5 (mod p). For example, if p = 7, then a = 3; and if p = 29, then a = 13. Let R be the ring of integers in Q(
(a) Prove the following ideal factorization holds in R:
(p) =
a +
− 5 , p
a −
− 5 , p
(b) What is the norm of the ideal
a +
− 5 , p
in R? Justify your answer.
(c) Is
a prime ideal? Is
a prime ideal? Justify your answers.
(d) Is
a principal ideal? Is
a principal ideal? Justify your answers.
(e) Is
a principal ideal? Justify your answer.
is odd should be easy.)