a) $ n^2004$ is congruent to $ 6 mod 9$ n obviously has to be relatively prime to 9, since otherwise $ n^2004$ is congruent to 0 mod 9, but then since phi(9) is 6, $ n^6$ is congruent to 1 mod 9, so $ n^2004$ is congruent to 1 mod 9, so no solution. b) $ n^2006$ is congruent to 8 mod 9 again, n is relatively prime, blah, blah, blah, $ n^2$ is congruent to 8 mod 9. But the quad. res. mod 9 are 1,4,0,7, contradiction!