- #1
Petr Mugver
- 279
- 0
For any natural number a and b, with b > 1, two natural numbers x and y can always be found such that b^x - b^y is divisible by a.
Until now, with many tries of a and b, I have always found corresponding x and y.
I would appreciate if somebody could give me a proof or a counter example.
Until now, with many tries of a and b, I have always found corresponding x and y.
I would appreciate if somebody could give me a proof or a counter example.