| 2pq+1 = 2p-1 | since 2p-1 is odd then the quotient must be even, hence the 2p |
| 2p(q-k)+2pk+1 = 2p-1 | subtracting k from q, we have to add 2pk to the expresion |
| 2p(2pk+1)(t)+(2pk+1) = 2p-1 | by choosing the correct k, (q-k) is divisible by (2pk+1)with a factor of (t) |
| (2pk+1)(2pt+1) = 2p-1 | factoring (2pk+1) |
| 4ktp2+2tp+2kp+1 = 2p-1 | multiplying the factors |
| 4ktp2+2p(t+k) = 2p-2 | subtracting 1 from each side and factoring 2p |
| 2ktp+(t+k) = (2p-2)/2p | dividing by 2p |
| 2ktp+t+k = (2)(2p-1-1)/2p | factoring out 2 |
| 2ktp+t+k = x | dividing by 2p gives us an integer x, via Little Fermat Theorem |
| (t)(2kp+1) = x-k | factoring t and subtracting k |
| t = (x-k)/(2kp+1) | dividing by (2kp+1) k > 0 |
This gives us a way to crank-out integer solutions for factors. The problem is the size of the numbers involved. Most solutions are very quick but 2101-1 has a k of 36,793,758,459 for one of the factors:
(2)(101)(36,793,758,459)+1