/*! 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 23 Let \((a, b, c)\) be a primitive... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 23

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 \(a^{2}+b^{2}=c^{2}\)

Short Answer

Expert verified
(a) Opposite parity as one cannot have both odd or even. (b) Derived from \\((c-a)(c+a) = b^2\\). (c) Opposite parity ensures correct factor distribution. (d) Using \\(p^2+q^2, p^2-q^2, 2pq\\) generates primitive triples.

Step by step solution

01

Understanding Pythagorean Triples

A Pythagorean triple \(a, b, c\) consists of three positive integers that satisfy the equation \(a^2 + b^2 = c^2\). A primitive Pythagorean triple means that the greatest common divisor (GCD) of the three numbers is 1.
02

Opposite Parity in Pythagorean Triples

Primitive Pythagorean triples have the property that one of \(a\) or \(b\) is odd and the other is even. This is because if both were odd, their squares would be congruent to 1 modulo 4, making \(a^2 + b^2\) congruent to 2 mod 4, which isn't possible since \(c^2\) must be congruent to 0 or 1 mod 4. Therefore, we assume \(a\) is odd and \(b\) is even.
03

Transforming the Equation

Since \(b\) is even, let \(b = 2m\). Substituting into \(a^2 + b^2 = c^2\) gives \(a^2 + (2m)^2 = c^2\). Rewriting it, we get \((c-a)(c+a) = (2m)^2\), which can be expressed as \((c-a)/2 \cdot (c+a)/2 = m^2\).
04

Ensuring Opposite Parity and HCF Condition

The factors \((c-a)/2\) and \((c+a)/2\) need to have an HCF of 1, and they must be of opposite parity. This ensures that the factors divide such that one is odd and another is even, thus mimicking the behaviour of the triple \(a, b, c\).
05

Expressing in Squares and Ensuring Parity

We can set \((c+a)/2 = p^2\) and \((c-a)/2 = q^2\) where \(p\) and \(q\) are integers of opposite parity. This is to satisfy the requirement that their sum and difference are both perfect squares. The condition \(HCF(p, q) = 1\) remains to ensure primitiveness.
06

Constructing the Triple

Given \(c + a = 2p^2\) and \(c - a = 2q^2\), solving for \(c\) and \(a\) gives \(c = p^2 + q^2\) and \(a = p^2 - q^2\). Since \(b = 2pq\) is an integer, \(a, b, c\) are integers forming a triangle where \(a^2 + b^2 = c^2\).
07

Verification for Primitive Triple

For any integers \(p\) and \(q\) such that they have opposite parity and \(HCF(p, q) = 1\), the triple \(a = p^2 - q^2, b = 2pq, c = p^2 + q^2\) satisfies \(a^2 + b^2 = c^2\) and is primitive.

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Key Concepts

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

Parity
The concept of parity in mathematics refers to whether an integer is odd or even. Parity plays a crucial role in understanding Pythagorean triples. In a primitive Pythagorean triple \(a, b, c\), one core property is that \(a\) and \(b\) have opposite parity. This means if \(a\) is odd, then \(b\) must be even and vice versa.
To understand why, consider the relationship \(a^2 + b^2 = c^2\). If both \(a\) and \(b\) were either odd or even, this equation would lead to contradictions in terms of modulo 4 arithmetic.
Thus, primitive Pythagorean triples maintain this opposite parity condition, which ensures the equation is satisfied correctly in integer form.
Greatest Common Divisor (GCD)
The Greatest Common Divisor (GCD) of a set of numbers is the largest positive integer that divides each of the numbers without leaving a remainder. In the context of primitive Pythagorean triples, the GCD of \(a, b, c\) must be 1.
This condition ensures that the triple is 'primitive', i.e., the numbers are coprime and not simply multiples of another Pythagorean triple.
By ensuring the GCD is 1, the three numbers \(a, b, c\) are the smallest set of numbers that fulfill the Pythagorean relationship \(a^2 + b^2 = c^2\) without any common factors.
Integer Solutions
Finding integer solutions to equations is an important aspect of number theory. In the case of Pythagorean triples, we are interested in solutions where \(a, b, c\) are all positive integers that satisfy \(a^2 + b^2 = c^2\).
These integer solutions demonstrate a connection between geometric concepts (representing the sides of a right triangle) and algebraic expressions.
By transforming variables, such as setting \(b = 2m\), and using perfect squares, we can generate solutions that maintain integer values while adhering to the Pythagorean equation.
Primitive Triples
A primitive triple is a set of three positive integers that are coprime (their GCD is 1), and they satisfy the Pythagorean theorem \(a^2 + b^2 = c^2\).
A primitive triple can be generated using integers \(p\) and \(q\) such that they are of opposite parity (one is odd, the other even) and \(GCD(p, q) = 1\).
The values of a primitive Pythagorean triple can be expressed in terms of \(p\) and \(q\):
  • \(a = p^2 - q^2\)
  • \(b = 2pq\)
  • \(c = p^2 + q^2\)

This expression guarantees all components of the triangle are integers, and ensures \(a, b, c\) form a valid primitive Pythagorean triple.

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Most popular questions from this chapter

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.

(a) Compute by mental arithmetic (using pencil only to record results), then learn by heart: (i) the squares of positive integers: first up to \(12^{2}\); then to \(31^{2}\) (ii) the cubes of positive integers up to \(11^{3}\) (iii) the powers of 2 up to \(2^{10}\). (b) How many squares are there: (i) \(<1000 ?\) (ii) \(<10000 ?\) (iii) \(<100000 ?\) (c) How many cubes are there: (i) (ii) (iii) \(<1000000 ?\) (d) (i) Which powers of 2 are squares? (ii) Which powers of 2 are cubes? (e) Find the smallest square greater than 1 that is also a cube. Find the next smallest. Evaluating powers, and the associated index laws, constitute an example of a direct operation. For each direct operation, we need to think carefully about the corresponding inverse operation - here "extracting roots". In particular, we need to be clear about the distinction between the fact that the equation \(x^{2}=4\) has two different solutions, while \(\sqrt{4}\) has just one value (namely 2).

(Pythagoras' Theorem) Let \(\triangle A B C\) be a right angled triangle, with a right angle at \(C .\) Draw the squares \(A C Q P, C B S R,\) and \(B A U T\) on the three sides, external to \(\triangle A B C\). Use the resulting diagram to prove in your head that the square \(B A U T\) on \(B A\) is equal to the sum of the other two squares by: \- drawing the line through \(C\) perpendicular to \(A B\), to meet \(A B\) at \(X\) and \(U T\) at \(Y\) \- observing that \(P A\) is parallel to \(Q C B\), so that \(\triangle A C P\) (half of the square \(A C Q P,\) with base \(A P\) and perpendicular height \(A C)\) is equal in area to \(\triangle A B P\) (with base \(A P\) and the same perpendicular height) \- noting that \(\triangle A B P\) is SAS-congruent to \(\triangle A U C,\) and that \(\triangle A U C\) is equal in area to \(\triangle A U X\) (half of rectangle \(A U Y X,\) with base \(A U\) and height \(A X)\) \- whence \(A C Q P\) is equal in area to rectangle \(A U Y X\) \- similarly \(B C R S\) is equal in area to \(B T Y X\). The proof in Problem 18 is the proof to be found in Euclid's Elements Book 1, Proposition 47. Unlike many proofs, \- it is clear what the proof depends on (namely SAS triangle congruence, and the area of a triangle), and \- it reveals exactly how the square on the hypotenuse \(A B\) divides into two summands \(-\) one equal to the square on \(A C\) and one equal to the square on \(B C\).

(a) Which of the prime numbers \(<100\) can be written as the sum of two squares? (b) Find an easy way to immediately write \(\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\) in the form \(\left(x^{2}+y^{2}\right)\). (This shows that the set of integers which can be written as the sum of two squares is "closed" under multiplication.) (c) Prove that no integer (and hence no prime number) of the form \(4 k+3\) can be written as the sum of two squares. (d) The only even prime number can clearly be written as a sum of two squares: \(2=1^{2}+1^{2}\). Euler \((1707-1783)\) proved that every odd prime number of the form \(4 k+1\) can be written as the sum of two squares in exactly one way. Find all integers \(<100\) that can be written as a sum of two squares. (e) For which integers \(N<100\) is it possible to construct a square of area \(N\), with vertices having integer coordinates? In Problem 25 parts (a) and (d) you had to decide which integers \(<100\) can be written as a sum of two squares as an exercise in mental arithmetic. In part (b) the fact that this set of integers is closed under multiplication turned out to be an application of the arithmetic of norms for complex numbers. Part (e) then interpreted sums of two squares geometrically by using Pythagoras' Theorem on the square lattice. These exercises are worth engaging in for their own sake. But it may also be of interest to know that writing an integer as a sum of two squares is a serious mathematical question \- and in more than one sense. Gauss \((1777-1855),\) in his book Disquisitiones arithmeticae (1801) gave a complete analysis of when an integer can be represented by a 'quadratic form', such as \(x^{2}+y^{2}\) (as in Problem 25) or \(x^{2}-2 y^{2}\) (as in Problem \(\mathbf{5 4}(\mathrm{c})\) in Chapter 2 ). A completely separate question (often attributed to Edward Waring \((1736-1798))\) concerns which integers can be expressed as a \(k^{\text {th }}\) power, or as a sum of \(n\) such powers. If we restrict to the case \(k=2\) (i.e. squares), then: \- When \(n=2,\) Euler \((1707-1783)\) proved that the integers that can be written as a sum of two squares are precisely those of the form $$ m^{2} \times p_{0} \times p_{1} \times p_{2} \times \cdots \times p_{s} $$ where \(p_{0}=1\) or \(2,\) and \(p_{1}

(a)(i) Explain why any integer that is a factor (or a divisor) of both \(m\) and \(n\) must also be a factor of their difference \(m-n,\) and of their sum \(m+n\). (ii) Prove that $$ H C F(m, n)=H C F(m-n, n) $$ (iii) Use this to calculate in your head \(H C F(1001,91)\) without factorising either number. (b)(i) Prove that: \(H C F(m, m+1)=1\). (ii) Find \(H C F(m, 2 m+1)\). (iii) Find \(H C F\left(m^{2}+1, m-1\right)\).
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