Question

Prove the following statements using any method from Chapters 4,5 or 6 . Explain why \(x^{2}+y^{2}-3=0\) not having any rational solutions (Exercise 20 ) implies \(x^{2}+y^{2}-3^{k}=0\) has no rational solutions for \(k\) an odd, positive integer.

Short answer

Having no rational solutions for the equation \(x^{2}+y^{2}-3=0\), infers that the equation \(x^{2}+y^{2}-3^{k}=0\) will also have no rational solutions when \(k\) is an odd, positive integer.

Step-by-step solution

Recalling the Property of the Given Equation

Recall that the equation \(x^{2}+y^{2}-3=0\) is telling us that the sum of squares of two rational numbers equals 3. By the properties of rational numbers, the sum of squares of two rational numbers must be a rational number. As 3 is a rational number, the condition for the equation to have a solution seems to be satisfied. But, in reality, it's impossible to find two rational squares that sum up to 3. This is due to number theoretical reasons related to the Pythagorean triples theorem. Hence, the equation \(x^{2}+y^{2}-3=0\) doesn't have any rational solutions.

Analyzing the New Equation

Now, analyze the new equation \(x^{2}+y^{2}-3^{k}=0\). Here, 3 is raised to the power of \(k\), where \(k\) is an odd, positive integer. It's a well-known property of exponentiation that the result of \(3^{k}\) will be a rational number if \(k\) is an odd, positive integer.

Using the Property of the Given Equation

Because \(3^{k}\) remains a rational number for positive odd integer values of \(k\), the same reasoning used for the equation \(x^{2}+y^{2}-3=0\) can be applied again to the equation \(x^{2}+y^{2}-3^{k}=0\). This equation is still representing a sum of two squares of some rational numbers that equals to a certain number \(3^{k}\). But again, it's impossible to find two rational squares that sum up to \(3^{k}\), for the same reasons mentioned above.

Conclusion

Thus, it can be concluded that if the equation \(x^{2}+y^{2}-3=0\) has no rational solutions, then the equation \(x^{2}+y^{2}-3^{k}=0\) also has no rational solutions for \(k\) an odd, positive integer. The original property of the equation persists, even when raising 3 to an odd, positive power

Key concepts

Pythagorean Triples Theorem

The Pythagorean triples theorem is a fundamental concept in number theory, especially when dealing with the sum of two squares, which is the crux of our problem involving the equation x^{2} + y^{2} - 3 = 0.

Pythagorean triples are sets of three positive integers, say (a, b, c), that satisfy the equation a^{2} + b^{2} = c^{2}. This theorem underpins the famous Pythagorean Theorem used in right-angled triangles. However, not every sum of two squares is a perfect square, and hence not every equation in the form x^{2} + y^{2} = z will have solutions that are rational numbers. For the specific case of the given equation, 3 is not the square of a rational number; therefore, there are no rational number solutions that would make the Pythagorean triples theorem hold true.

Properties of Rational Numbers

Understanding the properties of rational numbers is essential when analyzing equations like x^{2} + y^{2} - 3 = 0. A number is rational if it can be expressed as a fraction where both the numerator and the denominator are integers, and the denominator is not zero.

Rational numbers are closed under addition, subtraction, multiplication, and division (except by zero). This means that the sum, difference, product, or quotient of two rational numbers is also a rational number. However, the sum of the squares of two rational numbers does not always yield a rational number. The number 3 poses a unique challenge, as it can't be expressed as the sum of squares of two rational numbers. This property is pivotal in understanding why certain equations lack rational solutions.

Number Theory

Number theory is the study of numbers and the relationships between them. It deals with the properties and relationships of integers and is therefore intimately connected with our problem where we are examining whether certain equations have rational solutions.

In number theory, we learn that not all whole numbers are perfect squares and that it is impossible for certain integers, like 3, to be expressed as the sum of squares of two rational numbers. This challenge comes from the fact that there are no pairs of integers whose squares sum up to 3. By combining this knowledge with the properties of rational numbers, a deeper understanding of why certain equations remain unsolvable within rational numbers is achieved.

Exponentiation

Exponentiation is a mathematical operation, involving two numbers, the base and the exponent. When simplifying x^{2} + y^{2} - 3^{k} = 0, it is crucial to comprehend that 3^{k}, where k is an odd, positive integer, is still a rational number. This operation of raising a number to a power does not cause it to leave the set of rational numbers.

However, the key to understanding the equation's solutions—or lack thereof—lies not just in exponentiation but also in its interplay with the aforementioned concepts of Pythagorean triples and rational number properties. Regardless of the power to which 3 is raised, if the exponent is a positive integer, the result remains a rational number. But as 3 itself cannot be the sum of squares of two rational numbers, 3^{k} faces the same restraint, maintaining the equation's original property of having no rational solutions.