Problem: For every odd number p>1 we have:
Answer Choices:
A. (pβ1)21β(pβ1)β1 is divisible by pβ2
B. (pβ1)21β(pβ1)+1 is divisible by p
C. (pβ1)21β(pβ1) is divisible by p
D. (pβ1)21β(pβ1)+1 is divisible by p+1
E. (pβ1)21β(pβ1)β1 is divisible by pβ1
Solution:
Since p is odd and p>1, then 21β(pβ1)β₯1. In every case one factor of (pβ1)21β(pβ1)β1 will be [(pβ1)β1]=pβ2. The other choices are either possible only for special permissible values of p or not possible for any permissible values of p.