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 n4 will end with digit 1. 4n always end with a 4 as 4n = 4x4n-1, and 4n-1 = 16(n-1)/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.
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Posted by Bon
on 2004-08-12 14:46:53 |
Incomplete answer
