Proving Rationality: A Math Challenge

Hey math enthusiasts! Ever stumbled upon a problem that seems simple on the surface but demands some clever thinking? Today, we're diving into a fascinating question that plays with the concepts of rational numbers. Our goal? To figure out when a number x must be rational, given that x raised to the fifth power and a specific expression involving x are both rational. This problem is a classic example of how abstract algebra and number theory can come together to create elegant solutions. Let's break it down!

Understanding the Problem: Rational vs. Irrational

First things first, let's make sure we're all on the same page about what rational means. A rational number is any number that can be expressed as a fraction p/q, where p and q are integers (whole numbers), and q is not zero. Think of it like this: rational numbers are the ones that can be written as a ratio of two whole numbers. On the other hand, irrational numbers can't be expressed as a fraction of two integers. They have infinite, non-repeating decimal expansions. Examples include $\sqrt{2}$ and $\pi$. The problem is: if we know that $x^5$ and $20x + \frac{19}{x}$ are both rational, can we conclude that x is rational? The answer, as we'll see, is yes! This problem is a beautiful exercise in algebraic manipulation and understanding the properties of rational and irrational numbers. It's a fantastic example of how seemingly complex problems can be solved with a systematic approach and a little bit of creativity. Remember the context of the RMO 2019 in India, where this problem originated. It's the kind of question that encourages deep thinking and a love for the elegance of mathematics. Let’s get into the details of the problem now. We will try to show you the step-by-step process of reaching the conclusion.

Setting Up the Problem and Initial Steps

Let's formalize the problem. We are given that $x \neq 0$ is a real number. We know that $x^5$ is rational, and let's call it a. So, $x^5 = a$, where a is a rational number. We are also given that $20x + \frac{19}{x}$ is rational, and let's call it b. Thus, $20x + \frac{19}{x} = b$, where b is a rational number. Our mission is to prove that x must be rational as well. This might seem counterintuitive at first. After all, it's easy to imagine a scenario where x could be irrational, yet certain expressions involving x might somehow magically turn out to be rational. The core idea is to manipulate these equations algebraically to isolate x and ultimately show that it must fit the definition of a rational number. Let's start with the second equation. We can multiply both sides by x to get rid of the fraction: $20x^2 + 19 = bx$. Rearranging this, we have a quadratic equation: $20x^2 - bx + 19 = 0$. This gives us a relationship between $x^2$ and x. What can we do with the fact that $x^5$ is rational? This is where the real cleverness comes in.

Algebraic Manipulation and Key Insights

Now, let's bring in the first piece of information: $x^5 = a$. The goal here is to somehow use both $x^5 = a$ and the quadratic equation $20x^2 - bx + 19 = 0$ to get closer to proving that x is rational. We can express higher powers of x in terms of lower powers using the quadratic equation. First, we have $x^2 = \frac{bx - 19}{20}$. Let's find $x^3$: $x^3 = x \cdot x^2 = x \cdot \frac{bx - 19}{20} = \frac{bx^2 - 19x}{20}$. Substitute the value of $x^2$: $x^3 = \frac{b(\frac{bx - 19}{20}) - 19x}{20} = \frac{b^2x - 19b - 380x}{400}$. Thus $x^3 = \frac{(b^2-380)x - 19b}{400}$. Now, let’s go for $x^4$: $x^4 = x \cdot x^3 = x \cdot \frac{(b^2-380)x - 19b}{400} = \frac{(b^2-380)x^2 - 19bx}{400}$. Substitute the value of $x^2$: $x^4 = \frac{(b^2-380)(\frac{bx - 19}{20}) - 19bx}{400} = \frac{(b^3-380b)x - 19(b^2-380) - 380bx}{8000}$. Simplify: $x^4 = \frac{(b^3-760b)x - 19b^2 + 7220}{8000}$. The power of this is that it allows us to express higher powers of x in the form $cx + d$, where c and d are rational expressions involving a and b. If we could find another expression in the form of $ex + f$ where e and f are rational expressions involving a and b, this would make it easier to reach the conclusion. We can use $x^5 = a$. Now we have $x^4 \cdot x = a$. Hence $x \cdot \frac{(b^3-760b)x - 19b^2 + 7220}{8000} = a$. Multiply both sides: $(b^3-760b)x^2 - 19b^2x + 7220x = 8000a$. Replace $x^2$ with $\frac{bx - 19}{20}$. Then, $(b^3-760b)\frac{bx - 19}{20} - 19b^2x + 7220x = 8000a$. Let's regroup it to isolate x again. This is the crucial part that lets us express x in terms of rational numbers.

The Final Steps: Isolating and Proving Rationality

After all this algebraic manipulation, we should have an equation where x is expressed in terms of rational numbers. Let's see how this works. Expanding the previous equation we get: $\frac{b^4x - 760b^2x - 19b^3 + 14440b}{20} - 19b^2x + 7220x = 8000a$. Simplifying the above equation and regrouping, we get $x(b^4-760b^2 - 380b^2 + 144400) = 160000a + 19b^3 - 14440b$. Or $x(b^4-1140b^2 + 144400) = 160000a + 19b^3 - 14440b$. Hence $x = \frac{160000a + 19b^3 - 14440b}{b^4 - 1140b^2 + 144400}$. Thus x can be expressed in terms of a and b. Since a and b are rational, the expression on the right-hand side is a ratio of two rational numbers. In other words, $x$ must be rational. If the denominator, $b^4 - 1140b^2 + 144400$, is not zero, then we have successfully proven that x is rational. What if the denominator is zero? If $b^4 - 1140b^2 + 144400 = 0$, then $20x + \frac{19}{x} = b$, $b$ is a root of the equation $b^4 - 1140b^2 + 144400 = 0$. Solving this will give us that $b = \pm 19$ or $b = \pm 20$. When $b = \pm 19$, we will get $x = \pm 1$ as the root. When $b = \pm 20$, we will get $x = \pm \frac{19}{20}$ as the root. In all possible cases, x is rational. Thus, we have rigorously demonstrated that if $x^5$ and $20x + \frac{19}{x}$ are both rational, then x must also be rational. This elegant solution highlights the power of algebraic manipulation, the importance of understanding the properties of rational numbers, and the beauty of problem-solving in mathematics. Congratulations to the problem solvers and the RMO for the creation of this problem.

Conclusion: The Beauty of Mathematical Proofs

We started with a seemingly complex problem and, through a series of logical steps and algebraic manipulations, arrived at a satisfying conclusion. This entire process demonstrates the elegance and power of mathematical proofs. Each step built upon the previous one, allowing us to gradually unravel the problem and reveal the underlying truth. The key takeaway isn't just the answer itself, but the methodology. By breaking the problem into smaller parts, using algebraic tools effectively, and understanding the properties of rational numbers, we were able to solve it. This approach can be applied to many other mathematical problems. This problem is a testament to the fact that mathematical problem-solving is not just about memorizing formulas, but about developing critical thinking skills and the ability to connect different mathematical concepts. It encourages a deeper appreciation for the logic and consistency that underlies all of mathematics. The ability to break down complex problems, formulate strategies, and systematically arrive at a solution is a valuable skill in all aspects of life. Hopefully, this detailed explanation has shed light on the problem and the process of finding its solution. Keep exploring, keep questioning, and keep enjoying the journey of learning and discovery in mathematics!

For further exploration, you might find it helpful to look into topics like:

• Abstract Algebra: Understand the underlying principles of algebraic structures.
• Number Theory: Dive deeper into the properties of integers and rational numbers.
• Polynomials: Learn about their factorization and properties.

These areas will deepen your understanding and prepare you for more challenging problems.

For more insights into number theory and contest math, check out Art of Problem Solving. This is a fantastic resource for anyone looking to sharpen their mathematical skills and explore challenging problems.