The first expression is factorable into:
(n-1)(n^(n−2) + n^(n−3) + n^(n−4) + ... + n^2 + n + 1)
So dividing both sides by (n-1), the problem reduces to: (n-1) ≠ 0
(n^(n−2) + n^(n−3) + n^(n−4) + ... + n^2 + n + 1) is divisible by (n−1)
Each term of this reduced expression is 1 mod (n-1).
There are (n-1) terms on the LHS; added together, their sum is (n-1) mod (n-1).
This is equivalent to 0 mod (n-1).
So the reduced expression is divisible by (n−1).
So the original expression is divisible by (n−1)^2.
Provided n ≠ 1, which was given because n>1.
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Posted by Larry
on 2025-03-21 08:54:02 |
solution
