/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Reveals the triple (3,4,5) as th... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 22

Reveals the triple (3,4,5) as the first instance \((m=1)\) of a one-parameter infinite family of triples, which continues $$ (5,12,13)(m=2),(7,24,25)(m=3),(9,40,41)(m=4), \ldots $$ whose general term is $$ (2 m+1,2 m(m+1), 2 m(m+1)+1) $$ The triple (3,4,5) is also the first member of a quite different "one- parameter infinite family" of triples, which continues $$ (6,8,10),(9,12,15), \ldots $$ Here the triples are scaled-up versions of the first triple (3,4,5) . In general, common factors simply get in the way: If \(a^{2}+b^{2}=c^{2}\) and \(H C F(a, b)=s,\) then \(s^{2}\) divides \(a^{2}+b^{2},\) and \(a^{2}+b^{2}=c^{2} ;\) so \(s\) divides \(c .\) And if \(a^{2}+b^{2}=c^{2}\) and \(H C F(b, c)=s,\) then \(s^{2}\) divides \(c^{2}-b^{2}=a^{2},\) so \(s\) divides \(a .\) Hence a typical Pythagorean triple has the form \((s a, s b, s c)\) for some scale factor \(s,\) where \((a, b, c)\) is a triple of integers, no two of which have a common factor: any such triple is said to be primitive (that is, basic - like prime numbers). Every Pythagorean triple is an integer multiple of some primitive Pythagorean triple. The next problem invites you to find a simple formula for all primitive Pythagorean triples.

Short Answer

Expert verified
Primitive triples can be generated using \((m^2-n^2, 2mn, m^2+n^2)\) with appropriate \(m, n\).

Step by step solution

01

Understanding the problem

We know from the exercise that the first triple (3,4,5) is part of both an infinite sequence and scaled-up versions. For the infinite sequence, (3,4,5) is part of triples in the form \((2m+1, 2m(m+1), 2m(m+1)+1)\). For the scaled-up version, multiples of the first triple are formed, such as \((6,8,10)\).
02

Primitive Pythagorean triple formulation

A primitive Pythagorean triple can be generated using the formula \((a, b, c) = (m^2-n^2, 2mn, m^2+n^2)\), where \(m > n > 0\), \(m\) and \(n\) are coprime, and \(m - n\) is odd. This ensures that the triple is primitive since \(a\), \(b\), and \(c\) will have no common factor, other than 1.
03

Finding a simple formula

To find all primitive Pythagorean triples, we use the formula derived in the previous step. By ensuring that \(m\) and \(n\) meet the conditions 鈥� \(m > n > 0\), \(m\) and \(n\) are coprime, and \(m-n\) is odd 鈥� we can generate primitive triples. For example, if \(m = 2\) and \(n = 1\), \((a,b,c) = (3, 4, 5)\). This aligns with our earlier discussed sets, confirming the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Primitive Pythagorean Triples
Primitive Pythagorean Triples are a fascinating concept within number theory that refer to a set of three integers \((a, b, c)\) satisfying the equation \(a^2 + b^2 = c^2\), where the greatest common divisor of the three numbers is 1. This means that none of the numbers share a common factor other than 1. The first and most famous primitive Pythagorean triple is the set \((3, 4, 5)\).To generate new primitive triples, mathematicians use the formula \((m^2 - n^2, 2mn, m^2 + n^2)\), where \(m > n > 0\), \(m\) and \(n\) are coprime (they share no common factors other than 1), and \(m - n\) is odd. For example, using \(m=2\) and \(n=1\), we get the classic \((3, 4, 5)\). Using this formula ensures that the resulting triple has no factors in common, making it primitive.Primitive triples serve as the building blocks of all Pythagorean triples, as every triple can be scaled up from a primitive set by multiplying each member by the same number.
Coprime Numbers
Coprime numbers play a crucial role in creating primitive Pythagorean triples. Two integers are said to be coprime if their greatest common divisor is 1. This means they have no shared factors other than 1, making them fundamentally distinct numerically.In the context of primitive Pythagorean triples, the integers \(m\) and \(n\) used in the formula \((m^2 - n^2, 2mn, m^2 + n^2)\) must be coprime to ensure that the generated triple is primitive. The condition \(m - n\) being odd adds an additional layer of distinction, stabilizing the sequence to produce a valid triple each time.Seeking pairs of coprime numbers, particularly ones that also satisfy \(m - n\) being odd, becomes an intriguing mathematical exercise. This practice not only primes the field of number theory but also ensures a structured and logical pursuit of generating primitive triples.
Scaling in Mathematics
Scaling is a powerful concept in mathematics that allows us to resize objects or systems, making them larger or smaller. In the world of Pythagorean triples, scaling refers to generating new triples by multiplying an existing triple by the same integer.For instance, starting with the primitive triple \((3, 4, 5)\), one can generate the scaled version \((6, 8, 10)\) by multiplying each number in the triple by 2. This scaled triple still satisfies the Pythagorean theorem because \(6^2 + 8^2 = 10^2\). Scaling not only demonstrates the flexibility of mathematical structures but also reveals the infinite nature of triples. Each primitive Pythagorean triple can lead to infinitely many larger triples simply through multiplication by an integer. Thus, scaling serves as a bridge connecting the finite world of primitive triples to the infinite realm of all Pythagorean triples.
Mathematical Sequences
Mathematical sequences underpin the generation of Pythagorean triples, particularly through the use of parameterized expressions. A sequence is a set of numbers arranged in a specific order, usually defined by a formula or a rule.Within the study of Pythagorean triples, an example of such a sequence is given by \((2m+1, 2m(m+1), 2m(m+1)+1)\), where \(m\) is a parameter that can take on infinite positive integer values. This sequence, starting from \(m = 1\), produces triples like \((3, 4, 5)\), \((5, 12, 13)\), and so on.Sequences reveal the inherent patterns and regularities within triples. They allow us to often predict future triples and understand their behavior across different parameter sets. By examining such sequences, students gain insights into not only the mechanics but also the elegance of mathematical growth and structure.

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Most popular questions from this chapter

Problem \(29(A+B=C)\) The 3 by 1 rectangle \(A D E H\) consists of three adjacent unit squares: \(A B G H, B C F G, C D E F\) left to right, with \(A\) in the top left corner. Prove that $$ \angle D A E+\angle D B E=\angle D C E $$

(a) Factorise 12345 as a product of primes. (b) Using only mental arithmetic, make a list of all prime numbers up to 100 . (c)(i) Find a prime number which is one less than a square. (ii) Find another such prime. There are 4 prime numbers less than \(10 ; 25\) prime numbers less than \(100 ;\) and 168 prime numbers less than 1000 . Problem \(\mathbf{4}(\mathrm{c})\) is included to emphasise a frequently neglected message: Words and images are part of the way we communicate. But most of us cannot calculate with words and images. To make use of mathematics, we must routinely translate words into symbols. For example, unknown numbers need to be represented by symbols, and points in a geometric diagram need to be properly labelled, before we can begin to calculate, and to reason, effectively.

(a) In the "24 game" you are given four numbers. Your job is to use each number once, and to combine the four numbers using any three of the four basic arithmetical operations - using the same operation more than once if you wish, and as many brackets as you like (but never concatenating different numbers, such as "3" and "4" to make " \(34 "\) ). If the given numbers are \(3,3,4,4,\) then one immediately sees \(3 \times 4+3 \times 4=24 .\) With 3,3,5 , 5 it may take a little longer, but is still fairly straightforward. However, you may find it more challenging to make 24 in this way: (i) using the four numbers 3,3,6,6 (ii) using the four numbers 3,3,7,7 (iii) using the four numbers 3,3,8,8 . (b) Suppose we restrict the numbers to be used each time to "four \(4 \mathrm{~s}\) " \((4,4,4,4),\) and change the goal from "make \(24 ",\) to "make each answer from \(0-10\) using exactly four \(4 \mathrm{~s} "\). (i) Which of the numbers \(0-10\) cannot be made? (ii) What if one is allowed to use squaring and square roots as well as the four basic operations? What is the first inaccessible integer? Calculating by turning the handle deterministically (as with addition, or multiplication, or multiplying out brackets, or differentiating) is a valuable skill. But such direct procedures are usually only the beginning. Using mathematics and solving problems generally depend on the corresponding inverse procedures \(-\) where a certain amount of juggling and insight is needed in order to work backwards (as with subtraction, or division, or factorisation, or integration). For example, in applications of calculus, the main challenge is to solve differential equations (an inverse problem) rather than to differentiate known functions. Problem 14 captures the spirit of this idea in the simplest possible context of arithmetic: the required answer is given, and we have to find how (or whether) that answer can be generated. We will meet more interesting examples of this kind throughout the rest of the collection.

Let \((a, b, c)\) be a primitive Pythagorean triple. (a) Show that \(a\) and \(b\) have opposite parity (i.e. one is odd, the other even) so we may assume that \(a\) is odd and \(b\) is even. (b) Show that $$ \left(\frac{b}{2}\right)^{2}=\left(\frac{c-a}{2}\right)\left(\frac{c+a}{2}\right) $$ where $$ H C F\left(\frac{c-a}{2}, \frac{c+a}{2}\right)=1 $$ and \(\frac{c-a}{2}, \frac{c+a}{2}\) have opposite parity. (c) Conclude that $$ \frac{c+a}{2}=p^{2} \text { and } \frac{c-a}{2}=q^{2} $$ where \(H C F(p, q)=1\) and \(p\) and \(q\) have opposite parity, so that \(c=p^{2}+q^{2},\) \(a=p^{2}-q^{2}, b=2 p q\) (d) Check that any pair \(p, q\) having opposite parity and with \(H C F(p, q)=1\) gives rise to a primitive Pythagorean triple $$ c=p^{2}+q^{2}, \quad a=p^{2}-q^{2}, \quad b=2 p q $$ satisfying \(a^{2}+b^{2}=c^{2}\)

(a) Expand and simplify in your head: (i) \((\sqrt{2}+1)^{2}\) (ii) \((\sqrt{2}-1)^{2}\) (iii) \((1+\sqrt{2})^{3}\) (b) Simplify: (i) \(\sqrt{10+4 \sqrt{6}}\) (ii) \(\sqrt{5+2 \sqrt{6}}\) (iii) \(\sqrt{\frac{3+\sqrt{5}}{2}}\) (iv) \(\sqrt{10-2 \sqrt{5}}\) The expressions which occur in exercises to develop fluency in working with surds often appear arbitrary. But they may not be. The arithmetic of surds arises naturally: for example, some of the expressions in the previous problem have already featured in Problem \(\mathbf{3}(\mathrm{c}) .\) In particular, surds will feature whenever Pythagoras' Theorem is used to calculate lengths in geometry, or when a proportion arising from similar triangles requires us to solve a quadratic equation. So surd arithmetic is important. For example: \- A regular octagon with side length 1 can be surrounded by a square of side \(\sqrt{2}+1\) (which is also the diameter of its incircle); so the area of the regular octagon equals \((\sqrt{2}+1)^{2}-1\) (the square minus the four corners). \- \(\sqrt{2}-1\) features repeatedly in the attempt to apply the Euclidean algorithm, or anthyphairesis, to express \(\sqrt{2}\) as a "continued fraction". -\(\sqrt{10-2 \sqrt{5}}\) may look like an arbitrary, uninteresting repeated surd, but is in fact very interesting, and has already featured as \(4 \sin 36^{\circ}\) in Problem \(\mathbf{3}(\mathrm{c})\) \- One of the simplest ruler and compasses constructions for a regular pentagon \(A B C D E\) (see Problem 185) starts with a circle of radius 2 , centre \(O,\) and a point \(A\) on the circle, and in three steps constructs the next point \(B\) on the circle, where \(\underline{A B}\) is an edge of the inscribed regular pentagon, and $$ \underline{A B}=\sqrt{10-2 \sqrt{5}} $$
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