n^(n-1)-1 will invariably be divisible by (n-1), because (n-1) is a factor of that expression.
That division will leave an expression of (n-1) terms of form n^(n-1)+n^(n-2) +n^(n-3)+...1). Consider each of these mod(n-1).
(n^k-1) (for n>1, as required by the problem) is always divisible by (n-1), for the reason already given, so n^k is worth 1 mod(n-1). Thus each term of the expression will contribute 1, mod(n-1), for a total of (n-1) mod(n-1), or 0, thus ensuring divisibility by (n-1) for a second time.
Edited on March 21, 2025, 11:36 pm
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Posted by broll
on 2025-03-21 23:32:53 |
Possible Solution
