Number Theory Flashcards
Introduction to number theory (second year module) 5CCM224A
State bezouts lemma
Let a and b be integers (not both 0). Then there exists integers u and v such that gcd(a,b)=au+bv
What does it mean for two integers to be coprime?
the gcd of the two numbers is 1
define least common multiples
if a and b are integers then a common multiple of a and b is an integer c such that a|c and b|c. If a and b are both non-zero, the least common multiple of a and b is defined to be the smallest positive integer lcm(a,b) which is a common multiple of a and b.
What does b|a mean?
b divides a - a=qb for some q
define greatest common divisor
the largest divisor d of a and b such that d|a and d|b => gcd(a,b)=d
Euclidean algoritm
basically if a=qb+r then gcd(a,b)=gcd(b,r)
gcd(a,b)lcm(a,b)=?
|ab|
when does the linear diophantine equations ax+by=c have solutions?
only has solutions when c is a multiple of d=gcd(a,b)
How do you solve the linear diophantine equation ax+by=c?
(1) find x0, y0, gcd, using the euclidean algorithm a(x0)+b(y0)=d=gcd(a,b)
(2) the full set of solutions is x=x0+k.(b/g), y=y0+k(a/g) for all k∈ℤ
define a prime number
An integer p>1 is called a prime number if it has no positive divisors other than 1 and p.
fundamental theorem of arithmetic
every integer n>1 can be expressed uniquely (up to reordering) as a product of primes.
state euclids theorem and describe how the proof would go
there are infinitely many primes.
Proof by contradiction:
- assume there are finitely many primes
- there is an integer N such that N≥p for all primes p
- consider the integer n=N!+1=1x2x..x(N-1)xN +1
- by the fundamental theorem of arithmetic there is a prime p with p|n
- since p≤N we also have p|N!
- but 1=n-N! so that implies p|1 which is impossible
define congruence
let m be a non-zero integer, and let a,b∈ℤ. We say that a is congruent to b modulo m if m|(a-b).
define a multiplicative group
the multiplicative group of integers modulo m which is defined by:
ℤx(m)={ [a]m∈ℤ(m): a,m are coprime}
Hensels lemma
Let f(x)=a0+a1x+...+anx^n be a polynomial with integer coefficients, let p be a prime and let r be a positive integer. We let f'(x) be the derivative of f(x), so f'(x)=a1+2a2x+...+nanx^(n-1). suppose xr is an integer with f(xr)=0 (mod p^r) and f'(xr)≠0 (mod p) Then there exists x(r+1)∈ℤ satisfying f(x(r+1))=0 (mod p^(r+1)) and x(r+1)=xr (mod p^r) moreover the x(r+1) satisfying these properties is unique modulo p^(r+1) and we can take x(r+1)=xr-f(xr)u where u is an inverse of f'(xr) (modp).