Let's break down the problem and provide a step-by-step solution. **Problem Recap:** Let \( \sigma(n) \) denote the sum of the positive divisors of \( n \). Show that if \( k > 1 \) is a fixed integer, then the equation \( \sigma(n) = k \) has a **finite number** of solutions. --- ## Step 1: **Understanding \( \sigma(n) \)** - \( \sigma(n) \) is the sum of all positive divisors of \( n \). - For example, \( \sigma(6) = 1 + 2 + 3 + 6 = 12 \). --- ## Step 2: **Key Observation** - For \( n > 1 \), \( \sigma(n) \geq n + 1 \) since \( 1 \) and \( n \) are always divisors. - \( \sigma(n) \) is a strictly increasing function for powers of \( 2 \), i.e., \( \sigma(2^m) = 2^{m+1} - 1 \). --- ## Step 3: **Bounding \( n \) in terms of \( k \)** - If \( \sigma(n) = k \), then **certainly** \( n < k \) because \( \sigma(n) \geq n + 1 \) implies \[ k = \sigma(n) \geq n + 1 \implies n \leq k - 1. \] - So, any \( n \) satisfying \( \sigma(n) = k \) must have \( n \leq k - 1 \). --- ## Step 4: **Checking for Finiteness** - For a fixed \( k \), there are only finitely many integers \( n \) with \( n \leq k-1 \). - For each \( n \leq k-1 \), \( \sigma(n) \) can be computed and checked if it equals \( k \). --- ## Step 5: **Conclusion** Thus, the equation \( \sigma(n) = k \) has at most \( k-1 \) possible integer solutions for \( n \) (in fact, usually much fewer), so **the number of solutions is finite**. --- ## **Final Answer (Summary):** If \( k > 1 \) is a fixed integer, then the equation \( \sigma(n) = k \) has a finite number of solutions, because for any solution \( n \), we must have \( n < k \), so there are at most \( k-1 \) possibilities to check, and thus only finitely many solutions.