CLASS NUMBER DIVISIBILITY OF QUADRATIC FUNCTION
... The following lemma is due to Bae (the proof of Lemma 5.1 in [1] given there for real quadratic extension of k is easily seen to be valid for arbitrary quadratic extension of k). Lemma 2.1. Let F = k(y) be a quadratic extension of k, where y is a zero of x2 + Ax + B = 0 with (A, B) ∈ Ω. Let OF be th ...
... The following lemma is due to Bae (the proof of Lemma 5.1 in [1] given there for real quadratic extension of k is easily seen to be valid for arbitrary quadratic extension of k). Lemma 2.1. Let F = k(y) be a quadratic extension of k, where y is a zero of x2 + Ax + B = 0 with (A, B) ∈ Ω. Let OF be th ...
They are not equivalent
... covered in class and in discussion. If there is a topic for which no question is given below, you are still responsible for that topic. Also review the summaries at the end of Chapters 1 and 2. 1. State the contrapositive of the following: If x or y is even then x•y is even If xy is odd then x and y ...
... covered in class and in discussion. If there is a topic for which no question is given below, you are still responsible for that topic. Also review the summaries at the end of Chapters 1 and 2. 1. State the contrapositive of the following: If x or y is even then x•y is even If xy is odd then x and y ...
Unique factorization
... First of all, we analyzed the properties of hypothetical odd perfect numbers and showed that any odd perfect number W must exist in the form of (4k + 1) 4l + 1 M 2 , where (4k + 1, M ) = 1 , k , M Î N and 4k + 1 is prime . Using this result, we further studied the unit digits of 4k + 1 and M2 and su ...
... First of all, we analyzed the properties of hypothetical odd perfect numbers and showed that any odd perfect number W must exist in the form of (4k + 1) 4l + 1 M 2 , where (4k + 1, M ) = 1 , k , M Î N and 4k + 1 is prime . Using this result, we further studied the unit digits of 4k + 1 and M2 and su ...
