Find 3 different prime factors of 10^12 -1.
The Attempt at a Solution
I began trying to solve this with help from F.L.T. - if p is prime and p does NOT divide a, then a^(p-1) is congruent to 1 (mod p).
So I re-wrote, 10^(13-1) is congruent to 1 (mod13). Subtracting 1 from both sides of the congruence I have:
10^12 is congruent to 1 (mod13).
I believe 3 will be one prime factor since any multiple of 10 divided by 3 will yield 1 remainder. But I am stumped for others, and don't even know if finding 3 was systematically correct?
Now I also have a theorem that says I can divide both sides of the congruence under certain conditions, but I do not see that it will help? Any guidance, is FLT even a correct approach to this problem?