Question

Show that if \((u, v)\) is a solution to \(D x^{2}+2=y^{2}\), then \(\left(u_{1}, v_{1}\right)=\left(u v, v^{2}-1\right)\) is a solution to \(D x^{2}+1=y^{2} .\) Deduce a similar rule if \((u, v)\) is a solution to \(D x^{2}-2=y^{2}\).

Short answer

In summary, if we have an initial solution (u, v) to either of the equations Dx^2 + 2 = y^2 or Dx^2 - 2 = y^2, we can find a new solution by applying the following rules: 1. If (u, v) is a solution to Dx^2 + 2 = y^2, then (u₁, v₁) = (uv, v² - 1) is a solution to Dx^2 + 1 = y². 2. If (u, v) is a solution to Dx^2 - 2 = y^2, then (u₂, v₂) = (uv, v² + 1) is a solution to Dx² - 1 = y². These rules help us find new solutions to the given equations based on the initial solutions.

Step-by-step solution

Substituting (u,v) into the first equation

We are given that \((u, v)\) is a solution to the equation \(Dx^2 + 2 = y^2\). Therefore, \(Dx^2 + 2 = (uv)^2\).

Solve for D in terms of u and v

From the equation in step 1, we can solve for \(D\) by dividing both sides of the equation by \(x^2\). This gives us \(D = \frac{(uv)^2 - 2}{x^2}\).

Substituting (u1,v1) into the second equation

We are given that \((u_1, v_1)\) is related to \((u,v)\) by the equations \(u_1 = uv\), \(v_1 = v^2 - 1\). We need to show that \((u_1, v_1)\) is a solution to the equation \(Dx^2 + 1 = y^2\). Thus, let's substitute the expressions for \(u_1\) and \(v_1\) into this equation, which gives us \(\frac{(uv)^2 - 2}{x^2}x^2 + 1 = (v^2 - 1)^2\).

Simplify the equation

Now let's simplify the equation obtained in step 3. Since the term \(x^2\) cancels out in the left-hand side of the equation, we get \((uv)^2 - 1 = (v^2 - 1)^2\).

Prove that the second equation holds

To show that the second equation holds, we should expand the right-hand side of the equation and simplify it. So, \((v^2 - 1)^2 = v^4 - 2v^2 + 1\). Now let's substitute back \(u_1 = uv\) into the left-hand side of the equation, which gives us \((uv)^2 - 1 = u^2v^2 - 1\). It is now clear that both sides of the equation are equal, which means that \((u_1, v_1) = (uv, v^2 - 1)\) is indeed a solution to \(Dx^2 + 1 = y^2\).

Deduce a similar rule for the other case

When \((u, v)\) is a solution to the equation \(Dx^2 - 2 = y^2\), we can follow a similar approach to deduce a new rule. Let's assume that \((u_2, v_2)\) is a related solution to the equation \(Dx^2 - 1 = y^2\). By following the same steps as before, we can show that \((u_2, v_2) = (uv, v^2 + 1)\) is the rule for this case. Thus, the similar rule if \((u, v)\) is a solution to \(Dx^2 - 2 = y^2\) is that \((u_{2}, v_{2})=\left(u v, v^{2}+1\right)\) is a solution to \(D x^{2}-1=y^{2}\).

Key concepts

Mathematics History

The exploration of Diophantine equations, named after the ancient Greek mathematician Diophantus, dates back to a time when solutions often eluded finite computational techniques. Historically, Diophantine equations involve finding integer solutions to polynomial equations, serving as a challenging yet stimulating field in mathematics. The study of these equations laid the groundwork for many branches of mathematics, including number theory and algebra. The intrigue of finding whole number solutions to complex equations has propelled mathematical exploration for centuries. These problems connect modern mathematicians to their historical counterparts, offering a glimpse into the evolution of mathematical thought and techniques. By engaging with these ancient puzzles, mathematicians today continue to broaden the scope and application of mathematical research.

Equation Solutions

Diophantine equations are unique because they require solutions to be integers, unlike standard algebraic equations that can have any real number as a solution. Solving such equations isn't just about algebraic manipulation; it often involves insight and creativity. For instance, the equation given, \(D x^2 + 2 = y^2\), presents an interesting challenge of finding integer pairs \((u, v)\) that satisfy the condition. By finding one solution, we can often derive others through specific transformations or substitutions.

In our example, once \((u, v)\) is known, we can determine another solution to a slightly altered equation by applying a constructed transformation. The approach typically includes leveraging known solutions as a stepping stone to uncovering new ones. This method showcases the depth and beauty of the mathematical creative process.

Problem Solving Steps

Solving a Diophantine equation involves logical progression through sequenced steps.

• Identify Known Information: Start by understanding known solutions and given equations.
• Substitute and Manipulate: Introduce substitutions to explore variations of the base equation.
• Simplify Expressions: Break complex expressions down to clearer, more straightforward terms.
• Verify Solutions: Ensure that derived solutions satisfy all conditions of the equation set.

Following these steps systematically leads to revealing patterns or transformations between solutions, like transitioning from \((u, v)\) for one equation to \((u_1, v_1)\) for another. Always attack the problem by thinking critically about how each step affects the equation's integrity and outcome.

Algebraic Manipulation

Algebraic manipulation is a crucial skill in solving Diophantine equations, focusing on rearranging and simplifying equations to uncover solutions. When given equations like \(D x^2 + 2 = y^2\), a typical strategy involves substitution and simplification techniques. For instance, after substituting known values or expressions, probe into simplifying the resulting equation by combining like terms or canceling factors.

In our solution, manipulation is facilitated through replacing \(u_1 = uv\) and \(v_1 = v^2 - 1\) into \(D x^2 + 1 = y^2\) to confirm if the expression holds true across the board. This verification process involves expanding and rearranging expressions to confirm equality, showcasing algebra's power to bring clarity to integer solutions. With practice, these techniques refine one's ability to isolate variables and reveal true mathematical relationships hidden within the equations.