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Are there \(s, t \in \mathbb{Z}\) such that \(24 s+14 t=1\) ?

Short Answer

Expert verified
No, there are no such integers.

Step by step solution

01

Introduction to the Problem

We need to find integers \(s\) and \(t\) such that the equation \(24s + 14t = 1\) holds. This is a linear Diophantine equation of the form \(ax + by = c\), where \(a = 24\), \(b = 14\), and \(c = 1\). The equation has an integer solution if and only if the greatest common divisor (GCD) of \(a\) and \(b\) divides \(c\).
02

Calculate GCD of 24 and 14

Calculate the greatest common divisor (GCD) of 24 and 14 using the Euclidean algorithm:- Divide 24 by 14: 24 ÷ 14 = 1 remainder 10.- Divide 14 by 10: 14 ÷ 10 = 1 remainder 4.- Divide 10 by 4: 10 ÷ 4 = 2 remainder 2.- Divide 4 by 2: 4 ÷ 2 = 2 remainder 0.The last non-zero remainder is 2, so \(\text{GCD}(24, 14) = 2\).
03

GCD Check

Check if the GCD of 24 and 14 divides 1. Since 2 does not divide 1 (as 1 is not a multiple of 2), there are no integers \(s\) and \(t\) that satisfy the equation \(24s + 14t = 1\).
04

Conclusion

The initial condition for the existence of solutions to the Diophantine equation, that the GCD of 24 and 14 must divide 1, is not satisfied. Therefore, there are no integers \(s\) and \(t\) that satisfy the equation \(24s + 14t = 1\).

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Key Concepts

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

Euclidean Algorithm
The Euclidean Algorithm is a method for finding the greatest common divisor (GCD) of two integers. The GCD is the largest positive integer that divides both numbers without a remainder. The Euclidean Algorithm relies on the principle that the GCD of two numbers also divides their difference.
To use the Euclidean Algorithm, follow these steps:
  • Divide the larger number by the smaller one and consider the remainder.
  • Replace the larger number with the smaller number and the smaller number with the remainder.
  • Repeat the process until the remainder is zero.
  • The last non-zero remainder is the GCD of the two original numbers.
For example, to find the GCD of 24 and 14, we perform the following calculations:
  • Divide 24 by 14, remainder 10.
  • Divide 14 by 10, remainder 4.
  • Divide 10 by 4, remainder 2.
  • Divide 4 by 2, remainder 0.
Thus, the GCD of 24 and 14 is 2, as this was the last non-zero remainder.
Greatest Common Divisor
The Greatest Common Divisor (GCD) is an important concept in mathematics, especially when dealing with Diophantine equations like linear equations in integers. The GCD of two integers is the largest integer that can divide both numbers without leaving a remainder. The GCD is essential in determining the solvability of equations such as the one in the original exercise: 24s + 14t = 1.
Simply put:
  • If the GCD of two numbers divides a third number, then a linear equation involving these can potentially have integer solutions.
  • Otherwise, the equation has no integer solutions.
In the given problem, we found the GCD of 24 and 14 to be 2. If we want to solve the equation 24s + 14t = 1 for integers, 1 must be divisible by the GCD, which is not the case here. Therefore, no integers s and t can satisfy this equation.
Linear Equations in Integers
Linear equations in integers, often called linear Diophantine equations, are equations of the form ax + by = c where a, b, and c are integers, and x and y are variables constrained to be integers as well. These equations are named after the ancient mathematician Diophantus of Alexandria, who studied such equations extensively.
To determine if a linear Diophantine equation has solutions, follow these steps:
  • Calculate the greatest common divisor (GCD) of coefficients a and b.
  • Check if c is divisible by this GCD.
  • If so, the equation has integer solutions; otherwise, it does not.
In the case of the equation 24s + 14t = 1, the GCD of 24 and 14 is 2. Since 1 is not divisible by 2, the equation doesn't have any integer solutions. The rule here is that integer solutions exist if and only if the constant term (in this case, 1) is divisible by the GCD of the coefficients.

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Most popular questions from this chapter

Let \(x_{1}, x_{2}, \ldots\) be countably many indeterminates over \(\mathbb{Z}, R=\mathbb{Z}\left[x_{1}, x_{2}, \ldots\right]\), $$ Q_{i}=\left(\begin{array}{cc} 0 & 1 \\ 1 & x_{i} \end{array}\right) \in R^{2 \times 2} $$ for \(i \geq 1\), and \(R_{i}=Q_{i} \cdots Q_{1}\). We define the \(i\) th continuant polynomial \(c_{i} \in R\) recursively by \(c_{0}=0\), \(c_{1}=1\), and \(c_{i+1}=c_{i-1}+x_{i} c_{i}\) for \(i \geq 1\). Then \(c_{i} \in \mathbb{Z}\left[x_{1}, \ldots, x_{i-1}\right]\) for \(i \geq 1\). (i) List the first 10 continuant polynomials. (ii) Let \(T\) be the "shift homomorphism" \(T x_{i}=x_{i+1}\) for \(i \geq 1\). Show that \(c_{i+2}\left(0, x_{1}, x_{2}, \ldots, x_{i}\right)=T c_{i}\) for \(i \geq 0\). (iii) Show that \(R_{i}=\left(\begin{array}{cc}T c_{i-1} & c_{i} \\ T_{c_{i}} & c_{i+1}\end{array}\right)\) for \(i \geq 1\). (iv) Show that \(\operatorname{det} R_{i}=(-1)^{i}\), and conclude that \(\operatorname{gcd}\left(c_{i}, c_{i+1}\right)=1\) for \(i \geq 0\). (v) Let \(D\) be a Euclidean domain and \(r_{i}, q_{i}, s_{i}, t_{i} \in D\) for \(0 \leq i \leq \ell\) the results of the traditional Extended Euclidean Algorithm for \(r_{0}, r_{1}\). Show that $$ \begin{aligned} s_{i} &=c_{i-1}\left(-q_{2}, \ldots,-q_{i-1}\right)=(-1)^{i} c_{i-1}\left(q_{2}, \ldots, q_{i-1}\right) \\ t_{i} &=c_{i}\left(-q_{1}, \ldots,-q_{i-1}\right)=(-1)^{i-1} c_{i}\left(q_{1}, \ldots, q_{i-1}\right) \end{aligned} $$ for \(1 \leq i \leq \ell\) (vi) Write a MAPLE program that implements the traditional Extended Euclidean Algorithm and additionally computes all continuants \(c_{i}\left(q_{\ell-i+2}, \ldots, q_{\ell}\right)\) for \(r_{0}=x^{20}\) and \(r_{1}=x^{19}+2 x^{18}+x\) in \(\mathbb{Q}[x]\), where \(q_{1}, \ldots, q_{\ell}\) are the quotients in the traditional Extended Euclidean Algorithm.

Prove that \(\mathbb{Z}[x]\) is not a Euclidean domain. Hint: If it were, then we could compute \(s, t \in \mathbb{Z}[x]\) such that \(s \cdot 2+t \cdot x=\operatorname{gcd}(2, x)\), using the Extended Euclidean Algorithm.

Let \(F_{n}\) and \(F_{n+1}\) be consecutive terms in the Fibonacci sequence. Show that \(\operatorname{gcd}\left(F_{n+1}, F_{n}\right)=1\).

Show that \(\\{s f+t g: s, t \in \mathbb{Z}\\}=\\{k \cdot \operatorname{gcd}(f, g): k \in \mathbb{Z}\\}\) holds for all \(f, g \in \mathbb{Z}\). (In other words, the two ideals \(\langle f, g\rangle\) and \(\langle\operatorname{gcd}(f, g)\rangle\) are identical.)

Let \(R\) be a Euclidean domain, \(a, b, c \in R\), and \(\operatorname{gcd}(a, b)=1\). Prove the following: (i) \(a|b c \Longrightarrow a| c\), (ii) \(a \mid c\) and \(b|c \Longrightarrow a b| c\).

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