SOLUTION: show that the number {{{n^(n-1)-1}}} is divisible by {{{(n-1)^2}}} whenever {{{n>=2}}}

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Question 288851: show that the number n%5E%28n-1%29-1 is divisible by %28n-1%29%5E2 whenever n%3E=2
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
n%5E%28n-1%29-1 is divisible by %28n-1%29%5E2 whenever n%3E=2


Lemma: there exists non-negative integer q such that

n%5Ek%2F%28n-1%29=q%2B1%2F%28n-1%29

Proof:

By a factorization theorem n%5Ek-1 can be factored as:

n%5Ek-1=%28n-1%29%28n%5E%28k-1%29%2Bn%5E%28k-2%29%2B%22...%22%2Bn%5E2%2Bn%2B1%29

where there are k terms in the second parentheses

n%5Ek=%28n-1%29%28n%5E%28k-1%29%2Bn%5E%28k-2%29%2B%22...%22%2Bn%5E2%2Bn%2B1%29%2B1

Therefore n%5Ekmod%28n-1%29%22=%221

or there exists positive integer q such that

n%5Ek%2F%28n-1%29=q%2B1%2F%28n-1%29

Thus the lemma is proved.

n%5E%28n-1%29=%28n-1%29%28n%5E%28n-2%29%2Bn%5E%28n-3%29%2B%22...%22%2Bn%5E2%2Bn%2B1%29

where there are n-1 terms in the parentheses on the right.

Now we wish to show that

%28n%5E%28n-1%29-1%29%2F%28n-1%29%5E2 is a positive integer.  Factoring the numerator:





%28n%5E%28n-2%29%2Bn%5E%28n-3%29%2B%22...%22%2Bn%5E2%2Bn%2B1%29%2F%28n-1%29=%22%22



By the lemma, there exist q%5B1%5D,q%5B2%5D,...q%5Bn-1%5D-%22%22 so
that the preceding expression equals

q%5Bn-1%5D-%22%22%22%22%2B1%2F%28n-1%29%29%22%29=%22

%22%28%22q%5B1%5D%2Bq%5B2%5D%2Bq%5B3%5D%2B%22...%22%2Bq%5Bn-1%5D-%22%29%2B%221%2F%28n-1%29%2B1%2F%28n-1%29%2B%22...%22%2B1%2F%28n-1%29

since there are n-1 terms this becomes:

%22%28%22q%5B1%5D%2Bq%5B2%5D%2Bq%5B3%5D%2B%22...%22%2Bq%5Bn-1%5D-%22%29%2B%22%28n-1%29%2F%28n-1%29

or

q%5B1%5D%2Bq%5B2%5D%2Bq%5B3%5D%2B%22...%22%2Bq%5Bn-1%5D-%22%22%22%22%2B1,

which is a positive integer.

Edwin