Obviously if n is even, then the expression will be even, thus not a prime. So we'll only need to consider odd n's.
If n ends with a digit 1, 3, 7, or 9, then n^{4} will end with digit 1. 4^{n} always end with a 4 as 4^{n} = 4x4^{n1}, and 4^{n1} = 16^{(n1)/2} which always end with 6. So adding them together will end with digit 5, which is always divisible by 5.
Now the problem are those n ends with 5. I haven't figured that out yet. My observation is that it always end with a "49" with the 3rd and 4th last digit being ordered multiple of 8, starting from 16 (5), 24 (15), 32 (25), 40 (35), etc... For those curious,
5  1649
15  xxx2449
25  xxx3249
35  xxx4049
45  xxx4849
55  xxx5649
65  xxx6449
75  xxx7249
85  xxx8049
95  xxx8849
105 xxx9649
This can be proven similar to the way above, setting n = 10k+5.

Posted by Bon
on 20040812 14:46:53 