Provide valid reasoning for your answer.

m=2 is a trivial solution<o:p></o:p>

For m>2, m=integer, note that m and m-1 are always relatively prime.<o:p></o:p>

m^(m-1) = m * m * (m^(m-3)).  If this is to be divisible by (m-1)^2, then the last term must be divisible by (m-1)^2.  But the last term can then be written as m * m * (m^(m-5)).  Again the new last term must be devisable by (m-1)^2.  Follow this process to it’s logical end and you get two cases:<o:p></o:p>

1)      1) For m = odd, you get to m^(m-(m-2))= m^2, which is not divisible by (m-1)^2<o:p></o:p>

2)      2) For m = even, you get to m^1=m, which is also not divisible by (m-1)^2<o:p></o:p>

Therefore I believe m=2 is the only solution.<o:p></o:p>