First let's solve case with length 2. a*B + a = N <=> a*(B + 1) = N, it means that multipliers are divisors of N, so you need check all of them. Calculate primes up to sqrt(10^14), factorise N using them and recursively build all divisors, check them for correctness. Then solve case with length >= 4. It is easy to prove B in that case is < N^(1/3), so you can check all B up to it.