Show that if $m$ and $n$ are integers such that $(m,n)=1$ , then $(m+n,m-n)=1$ or 2 .

Step 1: Write down the given information and the formula for the GCD of two integers We are given that $(m,n)=1$ , where m and n are integers. We need to find $(m+n,m-n)$ . The GCD of two integers can be found using the Euclidean algorithm. Step 2: Apply the Euclidean algorithm to find the GCD of (m+n) and (m-n) Let's call the GCD of (m+n) and (m-n) 'd'. So, $d=(m+n,m-n)$ . According to the Euclidean algorithm, we need to find $(m-n,2n)$ . Since $(m,n)=1$ , we know that m and n have no common divisors other than 1. The GCD of (m-n) and 2n will give us an idea of the relation between (m+n, m-n) and 2. Step 3: Analyze the parity of m and n Since m and n are integers, they can either be odd or even. We'll consider the four possible cases: Case 1: m and n are both even. Case 2: m is even and n is odd. Case 3: m is odd and n is even. Case 4: m and n are both odd. Step 4: Determine the GCD of (m-n) and 2n in each case In each case, we'll find the GCD of (m-n) and 2n and describe the implications for the GCD of (m+n) and (m-n): Case 1: Both m and n are even. - In this case, (m-n) is even and 2n is also even. - Thus, $(m-n,2n)=2$ , which implies that $(m+n,m-n)=2$ . Case 2: m is even and n is odd. - In this case, (m-n) is odd and 2n is even. - Thus, $(m-n,2n)=1$ , which implies that $(m+n,m-n)=1$ . Case 3: m is odd and n is even. - In this case, similar to Case 2, (m-n) is odd and 2n is even. - Thus, $(m-n,2n)=1$ , which implies that $(m+n,m-n)=1$ . Case 4: Both m and n are odd. - In this case, (m-n) is even and 2n is even. - Thus, $(m-n,2n)=2$ , which implies that $(m+n,m-n)=2$ . Step 5: Conclude the result From the above cases, we have shown that the GCD of (m+n) and (m-n) can only be 1 or 2 for all possible pairs of integers m and n such that $(m,n)=1$ .