[seqfan] Re: 0, 2, 3, ?

[Jack Brennen]

There are no examples greater than 3.

First assume n >= 2.

Then note that n^n is congruent to 1^n modulo (n-1).

So n^n+1 is going to be 2 more than a multiple of (n-1).

The only way for n^n+1 to be divisible by n-1 is when 2 is divisible by 
n-1.  So n-1 is either 1 or 2; n is either 2 or 3.

   Jack




On 12/28/2014 4:36 PM, юрий герасимов wrote:
>
> Dear SeqFans,
> Numbers n such that n - 1 divides n^n + 1:  0,  2,  3, ... What in the 
> next ( > 3*10^5 ) one ?
> Best regards. JSG
>
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