Greatest common divisor of linear combination of two comprime numbers

Question

How to calculate $\gcd(2n+3m,n-m)$ if $\gcd(n,m)=1$ $\gcd(2n+3m,n-m)= \gcd(2n+3m+ 3(n-m),n-m)=\gcd(5n,n-m)= $ and i don't know. Plase help me

Answer 1

Suppose by symmetry $n \ge m$, then you have 2 side cases:

• $n = m \Rightarrow \gcd(2n+3m,n-m) = \gcd(2n+3m,0) = 2n+3m$
• $n = m + 1 \Rightarrow \gcd(2n+3m,n-m) = \gcd(2n+3m,1) = 2n+3(n-1) = 5n - 3$
• $n > m+1 \Rightarrow$ $$\gcd(2n+3m,n-m)= \gcd(2n+3m+ 3(n-m),n-m)=\gcd(5n,n-m)\\= \gcd(5n - 5\cdot (n-m), n-m) = \gcd(5m, n-m) \Rightarrow\\\gcd(5n,n-m) = \gcd(5m, n-m) = d \iff d \mid 5n, 5m, n-m$$ but since $$\gcd(n,m) =1$$ you must have $$\begin{cases} 5 \mid n-m \Rightarrow d = 5\\ 5 \nmid n-m \Rightarrow d = 1 \end{cases} \Rightarrow \gcd(2n+3m,n-m) = \gcd(5,n-m)$$