Mathematics Assignment: Number Theory

Running head: NUMBER THEORY 1
Mathematics Assignment: Number Theory
Student Name
Institution

NUMBER THEORY 2
Question 1
We prove the question by contradiction method
Suppose that u, v ∈ N are prime and both are not odd therefore one is even and the other is odd.
Let’s assume that:
i) U is an odd number and
ii) that (2 uv, u2−v2 ,u2 + v2) is primitive
Since (2 uv, u2−v2 ,u2 + v2) is primitive, there is a prime number , p which divides 2 uv,
u2−v2 ,∧u2 + v2.
If u2 is odd and v2 is even then u2 + v2 is odd. Consequently, p is not equal to 2 (p≠2)
According to the basic properties of divisors, the prime p divides:
(u¿¿ 2−v2 )+(u2 + v2) ¿ = 2u2 and (u¿¿ 2−v2 )−(u2+ v2)¿ = 2 v2
Additionally, since p≠2 the prime number, p divides u2∧v2 by default therefore the prime
number p divides u and v. However, u and v are prime so this is impossible. Therefore, ( 2 uv,
u2−v2 ,u2 + v2) is a primitive Pythagorean triple.
Question 2
For all integers x,y, x2, y2 ≥ 0; for nonzero x,y, we have the sharper inequality x2,y2 ≥ 1. In
General, if y≠0, then x2+2y2 ≥ 2y2≥ 2 > 1, then there's no solution with y≠0. Solving x2 = 1 over
the integers yields the obvious two solutions: (x,y) = (1,0) or (-1,0).
To begin, let's introduce some notation and relevant definitions.
i. Definition: Z[√2] := { u+v√2 : u,v in Z }.
ii. Definition: Let a,b in Z. Then the conjugate of u+v√2 in Z[√2], which I'll denote as
conj(u+v√2), is u-v√2.