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Running head: NUMBER THEORY 1
Mathematics Assignment: Number Theory
Student Name
Institution
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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 (,
) is primitive
Since (,
) is primitive, there is a prime number , p which divides ,

.
If
is odd and
is even then
is odd. Consequently, p is not equal to 2 (p2)
According to the basic properties of divisors, the prime p divides:
󰇛
󰇜 󰇛
󰇜 = 2
and 󰇛
󰇜 󰇛
󰇜 = 2
Additionally, since p2 the prime number, p divides

by default therefore the prime
number p divides u and v. However, u and v are prime so this is impossible. Therefore, (,
) is a primitive Pythagorean triple.
Question 2
For all integers x,y, x
2
, y
2
≥ 0; for nonzero x,y, we have the sharper inequality x
2
,y
2
≥ 1. In
General, if y≠0, then x
2
+2y
2
≥ 2y
2
2 > 1, then there's no solution with y≠0. Solving x
2
= 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.
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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 u
2
-
2v
2
. That is, the norm of an element in Z[√2] is the product of that element with its
conjugate.
Expressing
+ 
=1 in 󰇛 󰇜
󰇛 󰇜
= 1 while
- 
=1 in 󰇛
󰇜
󰇛 󰇜
= 1
Question 3
Following the claim is that if (m1, n1) and (m2, n2) are integer solutions to x
2
-2y
2
= 1, then you
can multiply the corresponding elements of Z[√2] to produce an integer solution to x
2
-
2y
2
= 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, a
2
-2b
2
= 1 if and only if a
2
-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 x
2
-2y
2
= 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[
]. We have N(3+2
) = 3
2
-2(2
2
) = 9-8 = 1, so (3,2) is a solution to the
Diophantine equation x
2
-2y
2
= 1. Moreover, it's not one of the trivial solutions (1,0) or (-1,0).
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NUMBER THEORY 4
Now: since α in Z[√2] produces an integer solution, the propositions above show that α
2
, α
3
, α
4
,
... α
n
are all solutions, too.
Furthermore, note that α > 1, so we have 1 < α < α
2
< α
3
< α
4
< ..., 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 x
2
-2y
2
= 1, so there are indeed infinitely such solutions as claimed.
As for approximations to √2, assume (x,y) is a solution to x
2
-2y
2
= 1; where without loss of
generality, x,y ≥ 0. Then
x
2
-2y
2
= 1
= 2 +
,iff y≠0
=
Thus
=
Provided |y| is large. Therefore, a solution (x,y) to x
2
-2y
2
= 1 produces a rational number
which
is a good rational approximation to
.,
As y gets larger and larger, our approximations to
gets more accurate
References

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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 (2𝑢𝑣, 𝑢2 − 𝑣 2 , 𝑢2 + 𝑣 2 ) is primitive Since (2𝑢𝑣, 𝑢2 − 𝑣 2 , 𝑢2 + 𝑣 2 ) is primitive, there is a prime number , p which divides 2𝑢𝑣, 𝑢2 − 𝑣 2 , and 𝑢2 + 𝑣 2 . If 𝑢2 is odd and 𝑣 2 is even then 𝑢2 + 𝑣 2 is odd. Consequently, p is not equal to 2 (p≠2) According to the basic properties of divisors, the prime p divides: (𝑢2 − 𝑣 2 ) + (𝑢2 + 𝑣 2 ) = 2𝑢2 and (𝑢2 − 𝑣 2 ) − (𝑢2 + 𝑣 2 ) = 2𝑣 2 Additionally, since p≠2 the prime number, p divides 𝑢2 and 𝑣 2 by default therefore the prime number p divides u and v. However, u and v are prime so this is impossible. Therefore, (2𝑢𝑣, 𝑢2 − 𝑣 2 , 𝑢2 + 𝑣 2 ) 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 iii. 3 Definition: Let u,v in Z. Then the norm of u+v√2 in Z[√2], denoted N(u+v√2), is u22v2. That is, the norm of an element in Z[√2] is the product of that element with its conjugate. Expressing 𝑋 2 + 2𝑌 2 =1 in (𝑢 + 𝑣√2)2 + 2(𝑢 − 𝑣√2)2 = 1 while 𝑋 2 - 2𝑌 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 𝑥 𝑦 1 = 2 + 𝑦 2 ,iff y≠0 1 = √2 + 𝑦 2 𝑥 Thus 𝑦 =√2 𝑥 Provided |y| is large. Therefore, a solution (x,y) to x2-2y2 = 1 produces a rational number 𝑦 which is a good rational approximation to√2., As y gets larger and larger, our approximations to √2 gets more accurate References Name: Description: ...
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