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Chapter 8: Problem 6

(a) Find integers \(u\) and \(v\) such that \(9 u+14 v=1\) or explain why it is not possible to do so. Then find integers \(x\) and \(y\) such that \(9 x+14 y=10\) or explain why it is not possible to do so. (b) Find integers \(x\) and \(y\) such that \(9 x+15 y=10\) or explain why it is not possible to do so. (c) Find integers \(x\) and \(y\) such that \(9 x+15 y=3162\) or explain why it is not possible to do so.

Short Answer

Expert verified
Short Answer: (a) Solutions: \(u=14k\) and \(v=1-9k\); \(x=140k\) and \(y=10-90k\). (b) There are no integer solutions for \(9x + 15y = 10\). (c) Solutions: \(x=-529 - 5k\) and \(y=1053 + 3k\).

Step by step solution

01

Part (a): Finding integer solutions for \(9u + 14v = 1\) and \(9x + 14y = 10\)

First, we'll find the greatest common divisor of 9 and 14 through Euclid's algorithm. \(9 = 14 \cdot 0 + 9\) \(14 = 9 \cdot 1 + 5\) \(9 = 5 \cdot 1 + 4\) \(5 = 4 \cdot 1 +1\) Since gcd(9,14) = 1, a solution exists for the equation \(9u + 14v = 1\). Now, we'll use the extended Euclidean algorithm to find a particular solution for integers \(u\) and \(v\). - Start with \(1 \cdot 9 + (-1) \cdot 14 = (-5)\) - Multiply by -1: \((-1) \cdot 9 + 1 \cdot 14 = 5\) - Add 1 multiplied by 9: \((-1+1) \cdot 9 + 1 \cdot 14 = 1\) - Particular solution: \(u_0 = 0\) and \(v_0 = 1\) The general solution is given by \(u = u_0 + \frac{14k}{\text{gcd}(9,14)}\) and \(v = v_0 - \frac{9k}{\text{gcd}(9,14)}\), with \(k \in \mathbb{Z}\). Here, \(u = 0 + 14k\) and \(v = 1 - 9k\) For the equation \(9x + 14y = 10\), since gcd(9,14) = 1 divides 10, a solution exists. We can modify the general solution we found before by multiplying both sides by 10. - Particular solution: \(u_1 = 0 \cdot 10\) and \(v_1 = 1 \cdot 10\) - The general solution is given by \(x = u_1 + \frac{14k}{\text{gcd}(9,14)}\) and \(y = v_1 - \frac{9k}{\text{gcd}(9,14)}\). - Here, \(x = 140k\) and \(y = 10 - 90k\).
02

Part (b): Finding integer solutions for \(9x + 15y = 10\)

For this part, we need to find the greatest common divisor of 9 and 15 through Euclid's algorithm. \(15 = 9 \cdot 1 + 6\) \(9 = 6 \cdot 1 + 3\) \(6 = 3 \cdot 2 + 0\) Since gcd(9,15) = 3, which doesn't divide 10, no integer solutions exist for the equation \(9x + 15y = 10\).
03

Part (c): Finding integer solutions for \(9x + 15y = 3162\)

For this part, we need to find the greatest common divisor of 9 and 15, which we already calculated above as 3. A solution exists since 3 divides 3162. Using the extended Euclidean algorithm for the equation \(9x + 15y = 3\): - Start with \(1 \cdot 9 + (-1) \cdot 15 = (-6)\) - Multiply by -1 to get \((-1) \cdot 9 + 1 \cdot 15 = 6\) - Divide by the gcd: \(\frac{(-1)}{3} \cdot 9 + \frac{1}{3} \cdot 15 = 2\) - Multiply by \(\frac{3162}{2}\): \((-529) \cdot 9 + 1053 \cdot 15 = 3162\) So, the particular solution is when \(x = -529\) and \(y = 1053\). We can obtain the general solution by adding or subtracting multiples of \(\frac{15}{\text{gcd}(9,15)} = 5\) and \(\frac{9}{\text{gcd}(9,15)} = 3\) to the particular solution \(x\) and \(y\), respectively, that is, \(x = -529 - 5k\) and \(y = 1053 + 3k\), with \(k \in \mathbb{Z}\)

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Key Concepts

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

integer solutions
Integer solutions refer to solutions of an equation where the variable values are whole numbers. In the given exercises, we are tasked to find such integer solutions for linear equations.
  • For instance, we need integers \(u\) and \(v\) such that \(9u + 14v = 1\).
  • Finding integer solutions is crucial as these equations often relate to practical scenarios requiring whole number values, like counting objects.
  • The presence of integer solutions is tied to the properties of coefficients in the equation.

Such problems are usually tackled using techniques like the Euclidean algorithm to find integer solutions, which reveals possible values for the variables that satisfy the equation. It's a key step in solving Diophantine equations.
Diophantine equations
Diophantine equations are equations where we look for integer solutions. These equations are named after the ancient Greek mathematician Diophantus and typically involve two or more unknowns and one resulting value or relation.
  • In the original exercise, we work with linear Diophantine equations such as \(9x + 14y = 10\).
  • These equations have a wide range of applications in number theory and cryptography.
  • The main principle is that we can find integer solutions, if and only if, the greatest common divisor (gcd) of the coefficients of \(x\) and \(y\) divides the constant term.

Often solved using the Euclidean algorithm and its extended version, Diophantine equations are a fundamental part of understanding integer relationships.
gcd (greatest common divisor)
The greatest common divisor (gcd) is a key concept in solving equations like those seen in the exercise. The gcd of two numbers is the largest positive integer that divides both numbers without leaving a remainder.
  • Understanding the gcd helps us determine if integer solutions exist for a Diophantine equation.
  • If the gcd of the coefficients in front of the variables divides the constant on the other side of the equality, an integer solution exists.
  • We use the Euclidean algorithm to compute the gcd, which is a systematic method of dividing and taking remainders.

The gcd is fundamental not only in solving linear Diophantine equations but also in simplifying fractions and finding least common multiples.

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Most popular questions from this chapter

(a) Let \(a\) and \(b\) be nonzero integers. If there exist integers \(x\) and \(y\) such that \(a x+b y=1,\) what conclusion can be made about \(\operatorname{gcd}(a, b) ?\) Explain. (b) Let \(a\) and \(b\) be nonzero integers. If there exist integers \(x\) and \(y\) such that \(a x+b y=2,\) what conclusion can be made about \(\operatorname{gcd}(a, b) ?\) Explain.

Prove the following proposition: Let \(n \in \mathbb{N}\). For each \(a \in \mathbb{Z},\) if \(\operatorname{gcd}(a, n)=1\), then for every \(b \in \mathbb{Z}\), there exists an \(x \in \mathbb{Z}\) such that \(a x \equiv b(\bmod n)\).

The goal of this exercise is to determine all (integer) solutions of the linear Diophantine equation in three variables \(12 x_{1}+9 x_{2}+16 x_{3}=20\) (a) First, notice that gcd \((12,9)=3\). Determine formulas that will generate all solutions for the linear Diophantine equation \(3 y+16 x_{3}=20\) (b) Explain why the solutions (for \(x_{1}\) and \(x_{2}\) ) of the Diophantine equation \(12 x_{1}+9 x_{2}=3 y\) can be used to generate solutions for \(12 x_{1}+9 x_{2}+16 x_{3}=20\) (c) Use the general value for \(y\) from Exercise (6a) to determine the solutions of \(12 x_{1}+9 x_{2}=3 y\) (d) Use the results from Exercises (6a) and (6c) to determine formulas that will generate all solutions for the Diophantine equation \(12 x_{1}+9 x_{2}+16 x_{3}=20\) Note: These formulas will involve two arbitrary integer parameters. Substitute specific values for these integers and then check the resulting solution in the original equation. Repeat this at least three times. (e) Check the general solution for \(12 x_{1}+9 x_{2}+16 x_{3}=20\) from Exercise (6d).

Prove the following proposition: For all natural numbers \(m\) and \(n,\) if \(m\) and \(n\) are twin primes other than the pair 3 and \(5,\) then 36 divides \(m n+1\) and \(m n+1\) is a perfect square.

The purpose of this exercise will be to prove that the nonlinear Diophantine equation \(3 x^{2}-y^{2}=-2\) has no solution. (a) Explain why if there is a solution of the Diophantine equation \(3 x^{2}-\) \(y^{2}=-2,\) then that solution must also be a solution of the congruence \(3 x^{2}-y^{2} \equiv-2(\bmod 3)\) (b) If there is a solution to the congruence \(3 x^{2}-y^{2} \equiv-2(\bmod 3)\), explain why there then must be an integer \(y\) such that \(y^{2} \equiv 2(\bmod 3)\). (c) Use a proof by contradiction to prove that the Diophantine equation \(3 x^{2}-y^{2}=-2\) has no solution.
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