91Ó°ÊÓ

When we speak about integer solutions, we mean finding values of variables that are whole numbers. In this context, our task is to find distinct integer values for the variables \( m \) and \( n \) such that \( \frac{1}{m} + \frac{1}{n} \) results in an integer. This involves some clever math analysis because we must satisfy the condition for it to turn out as an integer value.
In the problem at hand, we found that either \( m \) or \( n \) must equal 1. By setting one of these to 1, the equation simplifies significantly, making it easier to find other integer solutions. In our solution, when \( m = 1 \), \( n = 2 \) was one possible integer solution, and if \( n = 1 \), then \( m = 2 \) is another.

To solve the given problem, we first need to convert the fractions \( \frac{1}{m} \) and \( \frac{1}{n} \) to have a common denominator. The common denominator for these fractions is the product \( mn \).
By rewriting the fractions as \( \frac{n}{mn} + \frac{m}{mn} \), we can easily combine them into a single fraction: \( \frac{n+m}{mn} \). Using a common denominator allows us to work with fractions in a simplified way, facilitating their combination and making them easier to analyze further.

Equation analysis involves dissecting the equation to understand its components and behavior. In our exercise, the key was to make the resulting fraction \( \frac{n+m}{mn} \) equal to an integer \( k \).
This requires that the numerator, \( n+m \), must be divisible by the denominator, \( mn \). By equating \( \frac{n+m}{mn} = k \), and rearranging, we came to \( n+m = k(mn) \). This rearrangement is crucial as it sets the groundwork to find integer solutions and analyze conditions under which the equation holds true.

Distinct integers are different whole numbers. The term is used here to focus on finding different values for \( m \) and \( n \) such that they are not the same, thus excluding solutions like \( m = n \).
This adds an additional constraint to our problem. It forces us to carefully choose values so that \( m \) and \( n \) remain different yet ensure \( \frac{1}{m} + \frac{1}{n} \) still results in an integer. Our solution confirms that \( m = 1 \) and \( n = 2 \) or vice versa satisfies this condition, thereby achieving distinct integers.