Mathematics Assignment: Number Theory, Solutions, and Proofs

This document presents a comprehensive solution to a number theory assignment. The assignment explores concepts such as Pythagorean triples, Diophantine equations, and approximations of the square root of 2. Question 1 employs the contradiction method to prove a property of Pythagorean triples. Question 2 investigates integer solutions to a specific equation, introducing relevant definitions and notation. Question 3 builds on this by establishing propositions to demonstrate how multiplying solutions in Z[√2] generates new solutions to the equation. Finally, Question 4 uses a nontrivial solution to generate an infinite family of solutions and discusses how solutions provide rational approximations to the square root of 2. The document concludes with a reference to a textbook on elementary 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) ¿ = 2 u2 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.

NUMBER THEORY 3
iii. Definition: Let u,v in Z. Then the norm of u+v√2 in Z[√2], denoted N(u+v√2), is u2-
2v2. That is, the norm of an element in Z[√2] is the product of that element with its
conjugate.
Expressing X2+ 2 Y 2 =1 in (u+v √ 2)2+2(u−v √ 2)2= 1 while X2- 2 Y 2 =1 in
(u+v √ 2)2−2(u−v √ 2)2 = 1
Question 3
Following the claim is that if (m1, n1) and (m2, n2) are integer solutions to x2-2y2 = 1, then you
can multiply the corresponding elements of Z[√2] to produce an integer solution to x2-
2y2 = 1.
To solve this problem, three propositions are required :
i. Proposition 1: Let α in Z[√2], where α: = a+b√2 with a,b in Z. Then N(α) = 1 if and
only if N(conj α) = 1. That is, a2-2b2= 1 if and only if a2-2(-b)2 = 1.
ii. Proposition 2: Let α,β in Z[√2]. Then conj(αβ) = (conj α)(conj β).
iii. Proposition 3: Let α,β in Z[√2]. Then N(αβ) = N(α) N(β).
Let m1, n1, m2, n2 be integers,
And,
define α := m1 + n1√2, β = m2 + n2√2.
If N(α) = N(β) = 1, then N( (conj α)(conj β) ) = 1 hence (m,n)is also on the curve x2-2y2 = 1
Question 4
To show that there are infinitely many integer points on the curve, we will take a single
nontrivial solution, then use it to generate an infinite family of solutions. Consider α =
3+2√2 in Z[√2]. We have N(3+2√2) = 32-2(22) = 9-8 = 1, so (3,2) is a solution to the
Diophantine equation x2-2y2 = 1. Moreover, it's not one of the trivial solutions (1,0) or (-1,0).

NUMBER THEORY 4
Now: since α in Z[√2] produces an integer solution, the propositions above show that α2, α3,
α4, ... αn are all solutions, too.
Furthermore, note that α > 1, so we have 1 < α < α2 < α3 < α4 < ..., meaning these powers of α are
all distinct elements ofZ[√2]. That means these powers of α all yield distinct integer solutions to
the equation x2-2y2 = 1, so there are indeed infinitely such solutions as claimed.
As for approximations to √2, assume (x,y) is a solution to x2-2y2 = 1; where without loss of
generality, x,y ≥ 0. Then
x2-2y2 = 1
x2
y2 = 2 + 1
y2 ,iff y≠0
x
y = √2+ 1
y2
Thus x
y =√2
Provided |y| is large. Therefore, a solution (x,y) to x2-2y2 = 1 produces a rational number x
y which
is a good rational approximation to √2.,
As y gets larger and larger, our approximations to √2 gets more accurate

NUMBER THEORY 5
References
Burton, D. (2016). Elementary number theory. New Delhi, India: McGraw-Hill Education
(India) Private Limited.