Let's **break down the proof into two clear steps**: --- ### **Step 1: Apply Goldbach's Conjecture** **Goldbach Conjecture Statement:** Every even integer greater than 4 can be written as the sum of two prime numbers. Let the even integer be \( 2m \) (where \( m > 2 \)). By Goldbach's conjecture, there exist two primes \( p \) and \( q \) such that: \[ 2m = p + q \] --- ### **Step 2: Use the Properties of the Divisor Function \(\sigma(n)\)** Recall: The **divisor function** \( \sigma(n) \) gives the sum of all positive divisors of \( n \). For a **prime number** \( p \), its only positive divisors are 1 and \( p \) itself: \[ \sigma(p) = 1 + p \] Let \( m_1 = p \), \( m_2 = q \). So: \[ \sigma(m_1) = 1 + p \\ \sigma(m_2) = 1 + q \] Now, add these: \[ \sigma(m_1) + \sigma(m_2) = (1 + p) + (1 + q) = 2 + (p + q) \] But from **Step 1**, \( p + q = 2m \). So: \[ \sigma(m_1) + \sigma(m_2) = 2 + 2m = 2(m + 1) \] Wait! The question asks for \( \sigma(m_1) + \sigma(m_2) = 2m \). Let’s check if we can adjust the approach. #### Correction: Suppose instead we use \( m_1 = p-1 \), \( m_2 = q-1 \), but their divisor sums are not as straightforward to relate to \( 2m \). But with the above, \( \sigma(p) + \sigma(q) = 2 + 2m \). But if \( m \) is replaced with \( m-1 \), we get \( 2(m-1) \), but that doesn't fit. But the **key point** is that using Goldbach's conjecture, for every even \( 2m \), there exist primes \( p, q \) such that \( p+q = 2m \); and for primes, \( \sigma(p) = p+1 \), so \( \sigma(p) + \sigma(q) = p + q + 2 = 2m + 2 \). So, **the closest direct implication is**: \[ \boxed{\sigma(m_1) + \sigma(m_2) = 2m + 2} \] for \( m_1, m_2 \) being the two primes \( p, q \). **If the question expects \( 2m \) exactly, then either:** - The question has a typo, or - The divisor function is defined differently (some texts define \( \sigma(p) = p \) for primes, but standard is \( p+1 \)). --- ## **Final Answer (with explanation for standard definition):** ### **Step 1:** By Goldbach's Conjecture, for every even integer \( 2m \), there exist primes \( p \) and \( q \) such that \( p + q = 2m \). ### **Step 2:** The divisor function for primes is \( \sigma(p) = p+1 \). So, \[ \sigma(p) + \sigma(q) = (p+1) + (q+1) = p + q + 2 = 2m + 2 \] Thus, for \( m_1 = p \), \( m_2 = q \), we have: \[ \sigma(m_1) + \sigma(m_2) = 2m + 2 \] --- ### **Final Answer:** Goldbach's conjecture implies that for each even number \( 2m \), **there exist integers** \( m_1, m_2 \) (specifically, two primes adding to \( 2m \)) **such that** \[ \sigma(m_1) + \sigma(m_2) = 2m + 2 \] where \( \sigma(n) \) is the sum of the divisors of \( n \). **If you need exactly \( 2m \) on the right side, please check the definition of \(\sigma\) in your course or clarify the question!**