The Ramanujan-Nagell Theorem: Understanding the Proof 1
... First, we assume that n is even, in which case we have the factorization 7 = 2n − x2 = (2n/2 − x)(2n/2 + x). Clearly, both 2n/2 − x and 2n/2 + x must be integers. Because x is assumed positive and n > 0, then 2n/2 + x > 2n/2 − x, and we find that 7 = 2n/2 + x 1 = 2n/2 − x, from which we observe 8 = ...
... First, we assume that n is even, in which case we have the factorization 7 = 2n − x2 = (2n/2 − x)(2n/2 + x). Clearly, both 2n/2 − x and 2n/2 + x must be integers. Because x is assumed positive and n > 0, then 2n/2 + x > 2n/2 − x, and we find that 7 = 2n/2 + x 1 = 2n/2 − x, from which we observe 8 = ...
Math311W08Day3
... sequence. But we don’t know about the first N-1 terms. But there are only finitely many of these things, so let’s just say M = Max { a1 , a 2 , …, a n1 , a + 1} and ...
... sequence. But we don’t know about the first N-1 terms. But there are only finitely many of these things, so let’s just say M = Max { a1 , a 2 , …, a n1 , a + 1} and ...
Finite Abelian Groups as Galois Groups
... as h(X) = f (X)g(X), with f (X), g(X) monic polynomials in Q[X]. Then in fact, f (X) and g(X) have integer coefficients. To see this, write f (X) = ab f0 (X) and g(X) = dc g0 (X), where, f0 (X) and g0 (X) are primitive polynomials in Z[X]. Without loss of generality, assume the indicated fractions a ...
... as h(X) = f (X)g(X), with f (X), g(X) monic polynomials in Q[X]. Then in fact, f (X) and g(X) have integer coefficients. To see this, write f (X) = ab f0 (X) and g(X) = dc g0 (X), where, f0 (X) and g0 (X) are primitive polynomials in Z[X]. Without loss of generality, assume the indicated fractions a ...
Chapter 10
... **Note: The square root of any number yields both a + and a root since ++ = + and = !! Steps to Solving using Square Root Method 1. Get the x2 term alone on one side and the constant on the other. Remove any coefficient of x2 by division. 2. Take the square root of both sides remembering t ...
... **Note: The square root of any number yields both a + and a root since ++ = + and = !! Steps to Solving using Square Root Method 1. Get the x2 term alone on one side and the constant on the other. Remove any coefficient of x2 by division. 2. Take the square root of both sides remembering t ...
