Problem 19 - Algebra for JEE ADVANCED

Question

Let \(S(n)\) denote the number of ordered pairs \((x, y)\) satisfying \(1 / x\) \(+1 / y=1 / n\) where \(n>1\) and \(x, y, n \in N\) (i) Find the value of \(S(6)\). (ii) Show that if \(n\) is prime, then \(S(n)=3\) always.

Step-by-Step Solution

Answer

(i) \(S(6) = 9\). (ii) \(S(n) = 3\) for prime \(n\).

Step 1: Understanding the Equation

We start by rewriting the equation \(\frac{1}{x} + \frac{1}{y} = \frac{1}{n}\). By rewriting, we find a common denominator for the left side, leading to \(\frac{x + y}{xy} = \frac{1}{n}\). Cross-multiplying, we get \(n(x + y) = xy\). Rearranging, we obtain the quadratic equation \(xy - nx - ny = 0\).

Step 2: Factoring the Transformed Equation

To solve for \(x\) and \(y\), rearrange \(xy - x(n - y) = 0\) as \((x-n)(y-n)= n^2\). This implies \(x = n + d\) and \(y = n + \frac{n^2}{d}\), where \(d\) is a positive divisor of \(n^2\).

Step 3: Calculating S(6)

For \(n = 6\), the divisors of \(36\) (\(6^2\)) are \(1, 2, 3, 4, 6, 9, 12, 18, 36\). For each divisor \(d\), \(x = 6 + d\) and \(y = 6 + \frac{36}{d}\). Checking valid pairs of natural numbers, we count 9 solutions. Therefore, \(S(6) = 9\).

Step 4: Showing \(S(n) = 3\) for Prime \(n\)

If \(n\) is prime, the divisors of \(n^2\) are \(1, n, n^2\). For each divisor \(d\), valid solutions \((x, y)\) are: \((x, y) = (n+1, n + n^2), (n+n, n + \frac{n^2}{n}) = (n+n, n+n), (n+n^2, n+1)\). Thus, precisely 3 pairs satisfy the equation. Therefore, \(S(n) = 3\) when \(n\) is a prime number.

Key Concepts

Ordered Pairs

Ordered pairs are pairs of numbers where the order of the numbers matters. In mathematics, ordered pairs are used particularly in contexts involving coordinate geometry and solutions to equations.

• The notation \((x, y)\) signifies that \(x\) is related to \(y\) in a specific order.
• For the equation \(\frac{1}{x} + \frac{1}{y} = \frac{1}{n}\), finding ordered pairs means we look for values of \(x\) and \(y\) that satisfy this equation when \(n\) is given.

In the problem, the ordered pairs represent the solutions in terms of natural numbers to the given equation. These solutions are key to understanding how the equation can be satisfied under different circumstances (e.g., when \(n\) is a specific integer). It is particularly interesting to note the symmetry in many solutions where reversing \(x\) and \(y\) still satisfies the condition, hence making this an important factor in counting possible solutions.

Prime Numbers

Prime numbers are integers greater than 1 that have no divisors other than 1 and themselves. This property makes them fundamental in number theory.

• In our problem, if \(n\) is a prime number, it becomes crucial in determining the number of ordered pairs \((x, y)\).
• For a prime \(n\), the divisors of \(n^2\) are simply \(1\), \(n\), and \(n^2\) itself.

This characteristic is used in the solution to establish that for any prime number \(n\), there will always be exactly 3 solutions or ordered pairs. The uniqueness of primes simplifies the divisibility aspect of the equation, which in turn aids in reducing the complexity when seeking solutions.

Divisors

Divisors are numbers that divide a given number completely without leaving a remainder. Understanding divisors is central to breaking down numbers into manageable parts to solve equations like the one in our exercise.

• For example, with \(n = 6\), its square, \(36\), has divisors like \(1, 2, 3, \ldots, 36\).
• These divisors are used to find potential values for \(d\) in the factored equation \((x-n)(y-n) = n^2\).

By assigning each divisor \(d\), you can derive values for \(x\) and \(y\) as \(x = n + d\) and \(y = n + \frac{n^2}{d}\). In doing so, you identify which combinations of \(x\) and \(y\) satisfy our original equation, yielding valid pairs of natural numbers.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to make them easier to solve. This mathematical strategy is essential in arriving at solutions efficiently.

• The original equation \(\frac{1}{x} + \frac{1}{y} = \frac{1}{n}\) is transformed by algebraic manipulation into other forms like \(xy - nx - ny = 0\).
• Further, it is refactored into \((x-n)(y-n)=n^2\) to solve for specific values of \(x\) and \(y\).

This approach is particularly helpful as it reduces what starts as a fraction-based equation into a product of terms where divisors of \(n^2\) can directly lead to solutions. Such manipulation highlights the interconnectedness of algebraic techniques and their practical application in problem-solving.