Solving The Equation (x^2 - 7x + 11)(x^2 - 13x + 42) = 1 Finding Integer Solutions

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Introduction to the Problem

In this article, we delve into a fascinating problem from the realm of number theory, specifically focusing on finding the number of distinct positive integer-valued solutions to the equation extbf{(x2βˆ’7x+11)(x2βˆ’13x+42)=1(x^2 - 7x + 11)(x^2 - 13x + 42) = 1}. This problem requires a blend of algebraic manipulation, logical reasoning, and a keen eye for detail. The equation, seemingly simple at first glance, holds intricacies that demand a methodical approach to unravel its solutions. We will explore the factors of the equation, analyze the quadratic expressions, and determine the values of extit{x} that satisfy the given condition. The journey to the solution will not only involve mathematical techniques but also an understanding of the properties of integers and quadratic equations. By the end of this discussion, you will gain a comprehensive understanding of how to tackle such problems and appreciate the beauty of mathematical problem-solving. This problem is not just about finding the correct answer; it’s about understanding the underlying concepts and developing a strategy that can be applied to similar problems in the future. The challenge lies in identifying the possible integer solutions, which requires a systematic evaluation of the factors and the corresponding quadratic expressions. Let’s embark on this mathematical journey, step by step, to uncover the distinct positive integer solutions to this equation. Through careful analysis and logical deduction, we will navigate the complexities of the equation and arrive at a definitive answer.

Understanding the Equation

To effectively solve the equation extbf{(x2βˆ’7x+11)(x2βˆ’13x+42)=1(x^2 - 7x + 11)(x^2 - 13x + 42) = 1}, we must first understand its structure. The equation is a product of two quadratic expressions equaling 1. In the realm of integers, this presents a limited number of possibilities. For the product of two integers to be 1, either both integers must be 1, or both integers must be -1. This is because 1 multiplied by 1 equals 1, and -1 multiplied by -1 also equals 1. There are no other integer pairs whose product results in 1. Therefore, we can break down the problem into two distinct cases, each representing a set of equations to solve. These cases are:

  1. Both quadratic expressions equal 1:

    • x2βˆ’7x+11=1x^2 - 7x + 11 = 1
    • x2βˆ’13x+42=1x^2 - 13x + 42 = 1

  2. Both quadratic expressions equal -1:

    • x2βˆ’7x+11=βˆ’1x^2 - 7x + 11 = -1
    • x2βˆ’13x+42=βˆ’1x^2 - 13x + 42 = -1

By considering these cases separately, we can transform the original problem into a series of simpler quadratic equations. Solving each quadratic equation will give us potential values for extit{x}, which we must then validate against the original equation to ensure they are indeed solutions. This methodical approach of breaking down a complex problem into smaller, manageable parts is a fundamental strategy in problem-solving. It allows us to focus on specific aspects of the problem without being overwhelmed by the whole. In the following sections, we will delve into each of these cases, solving the quadratic equations and identifying the integer solutions that satisfy the original equation. Understanding this breakdown is crucial for navigating the complexities of the problem and arriving at the correct answer. The key is to remember that integer solutions are our primary focus, which significantly narrows down the possibilities we need to consider.

Case 1: Both Quadratic Expressions Equal 1

Let's begin by exploring the first case, where both quadratic expressions in the equation extbf{(x2βˆ’7x+11)(x2βˆ’13x+42)=1(x^2 - 7x + 11)(x^2 - 13x + 42) = 1} are equal to 1. This scenario gives us two separate quadratic equations to solve:

  1. x2βˆ’7x+11=1x^2 - 7x + 11 = 1
  2. x2βˆ’13x+42=1x^2 - 13x + 42 = 1

First, we'll tackle the equation x2βˆ’7x+11=1x^2 - 7x + 11 = 1. By subtracting 1 from both sides, we can rewrite it as x2βˆ’7x+10=0x^2 - 7x + 10 = 0. This quadratic equation can be factored into (xβˆ’2)(xβˆ’5)=0(x - 2)(x - 5) = 0. From this, we find two potential solutions: extit{x} = 2 and extit{x} = 5. These are the values of extit{x} that make the first quadratic expression equal to 1. Next, we move on to the second equation, x2βˆ’13x+42=1x^2 - 13x + 42 = 1. Subtracting 1 from both sides gives us x2βˆ’13x+41=0x^2 - 13x + 41 = 0. This quadratic equation does not factor easily, so we can use the quadratic formula to find its roots. The quadratic formula is given by x = rac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}, where extit{a}, extit{b}, and extit{c} are the coefficients of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. In this case, extit{a} = 1, extit{b} = -13, and extit{c} = 41. Plugging these values into the quadratic formula, we get:

x = rac{13 \\pm \\sqrt{(-13)^2 - 4(1)(41)}}{2(1)} = rac{13 \\pm \\sqrt{169 - 164}}{2} = rac{13 \\pm \\sqrt{5}}{2}

The solutions obtained are x = rac{13 + \\sqrt{5}}{2} and x = rac{13 - \\sqrt{5}}{2}. These values are not integers, and since we are looking for positive integer solutions, they do not satisfy our requirements. Therefore, from this part of Case 1, we have two potential integer solutions, extit{x} = 2 and extit{x} = 5, which we need to verify against the second original quadratic equation. In the next step, we will check if these values also satisfy the second equation, x2βˆ’13x+42=1x^2 - 13x + 42 = 1, to confirm whether they are valid solutions for the original problem.

Case 2: Both Quadratic Expressions Equal -1

Now, let's consider the second case where both quadratic expressions in the equation extbf{(x2βˆ’7x+11)(x2βˆ’13x+42)=1(x^2 - 7x + 11)(x^2 - 13x + 42) = 1} are equal to -1. This gives us the following two quadratic equations:

  1. x2βˆ’7x+11=βˆ’1x^2 - 7x + 11 = -1
  2. x2βˆ’13x+42=βˆ’1x^2 - 13x + 42 = -1

We begin with the equation x2βˆ’7x+11=βˆ’1x^2 - 7x + 11 = -1. Adding 1 to both sides, we get x2βˆ’7x+12=0x^2 - 7x + 12 = 0. This quadratic equation can be factored into (xβˆ’3)(xβˆ’4)=0(x - 3)(x - 4) = 0. Thus, the potential solutions are extit{x} = 3 and extit{x} = 4. These are the values of extit{x} that make the first quadratic expression equal to -1. Next, we consider the second equation, x2βˆ’13x+42=βˆ’1x^2 - 13x + 42 = -1. Adding 1 to both sides, we obtain x2βˆ’13x+43=0x^2 - 13x + 43 = 0. This quadratic equation does not factor easily, so we will use the quadratic formula to find its roots. Recall that the quadratic formula is x = rac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}, where extit{a}, extit{b}, and extit{c} are the coefficients of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. In this case, extit{a} = 1, extit{b} = -13, and extit{c} = 43. Plugging these values into the quadratic formula, we get:

x = rac{13 \\pm \\sqrt{(-13)^2 - 4(1)(43)}}{2(1)} = rac{13 \\pm \\sqrt{169 - 172}}{2} = rac{13 \\pm \\sqrt{-3}}{2}

The solutions obtained are x = rac{13 + \\sqrt{-3}}{2} and x = rac{13 - \\sqrt{-3}}{2}. These values involve the square root of a negative number, which means they are complex numbers and not integers. Since we are looking for positive integer solutions, these values do not satisfy our requirements. Therefore, from this part of Case 2, we have two potential integer solutions, extit{x} = 3 and extit{x} = 4, which we need to verify against the second original quadratic equation. In the following step, we will check if these values also satisfy the second equation, x2βˆ’13x+42=βˆ’1x^2 - 13x + 42 = -1, to confirm whether they are valid solutions for the original problem. This verification process is crucial to ensure that the solutions we find are consistent across both equations and, thus, are valid solutions for the original problem.

Verifying the Solutions

After analyzing both cases, we have identified four potential integer solutions: extit{x} = 2, extit{x} = 5 from Case 1, and extit{x} = 3, extit{x} = 4 from Case 2. To confirm whether these are indeed solutions to the original equation extbf{(x2βˆ’7x+11)(x2βˆ’13x+42)=1(x^2 - 7x + 11)(x^2 - 13x + 42) = 1}, we need to substitute each value back into the original equation and check if the equality holds. This verification step is essential because it ensures that the solutions we found satisfy the initial conditions of the problem. Without verification, we cannot be certain that our solutions are correct.

Let's start with extit{x} = 2:

  • (22βˆ’7(2)+11)(22βˆ’13(2)+42)=(4βˆ’14+11)(4βˆ’26+42)=(1)(20)=20(2^2 - 7(2) + 11)(2^2 - 13(2) + 42) = (4 - 14 + 11)(4 - 26 + 42) = (1)(20) = 20. This does not equal 1, so extit{x} = 2 is not a solution.

Next, we check extit{x} = 5:

  • (52βˆ’7(5)+11)(52βˆ’13(5)+42)=(25βˆ’35+11)(25βˆ’65+42)=(1)(2)=2(5^2 - 7(5) + 11)(5^2 - 13(5) + 42) = (25 - 35 + 11)(25 - 65 + 42) = (1)(2) = 2. This also does not equal 1, so extit{x} = 5 is not a solution.

Now, let's verify extit{x} = 3:

  • (32βˆ’7(3)+11)(32βˆ’13(3)+42)=(9βˆ’21+11)(9βˆ’39+42)=(βˆ’1)(12)=βˆ’12(3^2 - 7(3) + 11)(3^2 - 13(3) + 42) = (9 - 21 + 11)(9 - 39 + 42) = (-1)(12) = -12. This does not equal 1, so extit{x} = 3 is not a solution.

Finally, we check extit{x} = 4:

  • (42βˆ’7(4)+11)(42βˆ’13(4)+42)=(16βˆ’28+11)(16βˆ’52+42)=(βˆ’1)(6)=βˆ’6(4^2 - 7(4) + 11)(4^2 - 13(4) + 42) = (16 - 28 + 11)(16 - 52 + 42) = (-1)(6) = -6. This also does not equal 1, so extit{x} = 4 is not a solution.

After substituting each potential solution into the original equation, we find that none of them satisfy the equation. This means that there are no distinct positive integer-valued solutions to the equation (x2βˆ’7x+11)(x2βˆ’13x+42)=1(x^2 - 7x + 11)(x^2 - 13x + 42) = 1. This conclusion highlights the importance of verifying solutions, as the algebraic manipulations alone do not guarantee that the values obtained are correct in the context of the original problem. In this case, despite finding potential integer values, none of them held true when substituted back into the equation.

Conclusion

In conclusion, after a thorough analysis and verification process, we have determined that there are extbf{no distinct positive integer-valued solutions} to the equation extbf{(x2βˆ’7x+11)(x2βˆ’13x+42)=1(x^2 - 7x + 11)(x^2 - 13x + 42) = 1}. This problem showcased the importance of breaking down a complex equation into manageable cases, solving quadratic equations, and, most crucially, verifying the solutions. We explored two primary cases: one where both quadratic expressions equal 1, and another where both expressions equal -1. While we found potential integer solutions in each case, none of them satisfied the original equation upon verification.

This problem is a valuable exercise in mathematical problem-solving, highlighting the significance of a systematic approach. It demonstrates that finding potential solutions is only part of the process; the verification step is equally important to ensure the accuracy and validity of the results. The initial factorization and the use of the quadratic formula were key techniques employed in this solution. However, the final verification step revealed that these techniques, while useful, do not guarantee a correct answer without confirming the solutions in the original context. The absence of positive integer solutions underscores the subtlety of number theory problems, where seemingly straightforward equations can lead to unexpected outcomes. The process of elimination and careful verification is a fundamental skill in mathematics, and this problem serves as an excellent example of its application. By following a logical progression and verifying each step, we can confidently conclude that there are no distinct positive integer solutions to the given equation.