Positive integers a,b are such that 137 divides a+139b and 139 divides a+137b. Find the minimum possible value of a+b.
137 and 139 are consecutive odd numbers, so I will generalize the problem so that 137=2k-1 and 139=2k+1. The original problem is when k=69
Then 2k-1 divides a+(2k+1)*b and 2k+1 divides a+(2k-1)*b.
Looking at the specific answer in earlier comments, let a=4k-1 and b=k-1.
Then [a+(2k+1)*b]/(2k-1) = k-2 and [a+(2k-1)*b]/(2k-1) = k.
Then a+b=(4k-1)+(k-1) = 5k-2. So with the specific value k=69 then 5*69-2 = 343.
Generalization
