Using what we did in the previous problem, and knowing order = number of elements, I figured 25=$ 5^2 $
Thus, we need an $ f $(x) of degree 2 that's irreducible over $ Z_5 $[x].
I found $ f $(x) = $ x^2 + 2x + 3 $ to have no root in $ Z_5 $, thus $ Z_5[x] / f(x) $ would have $ 5^2 $ = 25 elements, thus a field of order 25! --Bcaulkin 21:57, 8 April 2009 (UTC)
