[seqfan] Re: Definition of primitive root (was Re: changes to A205989)

[David Wilson]

On 2/18/2012 8:45 PM, franktaw at netscape.net wrote:
> If you take that definition literally, *every* residue of any integer 
> that is not a unit modulo that integer is a primitive root. This is 
> obviously nonsense.
>
> You need to add that 10^(p-1) == 1 (mod p). At which point 10 is not a 
> primitive root of 2.
>

Alois: I apologize. My description of how to determine if 10 is a 
primitive root of prime p was incomplete. I neglected to mention that
10^(p-1) == 1 (mod p) is a necessary condition because I thought it was 
understood. I see you have updated your program to test this condition.

I can now see that it was my own inadequate instructions that led to 
this confusion. So again, please accept my apologies and thanks for your 
help with this sequence.

And thanks, Franklin, for rooting out the problem.

- Dave Wilson

> Alonso is correct; 10 is not a primitive root of 2.
>
> Franklin T. Adams-Watters
>
> -----Original Message-----
> From: Heinz, Alois <alois.heinz at hs-heilbronn.de>
>
> In "pink boxes" of A205989 ...
>
> Alonso del Arte asked: "What is the best way to figure out if a given
> prime has primitive root 10?"
>
> David W. Wilson answered: "The best way I know of is to verify that
> 10^((p-1)/q) != 1 (mod p) for each prime q dividing p-1."
>
> Now take p=2 => p-1=1, and NO prime q divides 1, thus ...
> 10^((p-1)/q) != 1 (mod p) for each prime q dividing p-1.
>
> So according to this algorithm 10 is a primitive root of 2.
>
> Alois
>
> Am 18.02.2012 23:39, schrieb David Wilson:
>> To clarify this to other readers:
>>
>> A205989(n) gives the smallest prime p >= 10^n with primitive root 10.
>> The question is about the value of a(0). Alois contends that 10 is a
>> primitive root modulo 2, in which case A205989(0) = 2. I contend that
>> 10 is not a primitive root modulo 2, which implies A205989(0) = 7
>> (since we would agree that 10 is not a primitive root modulo 3 or 5).
>> [ ... ]
>
>
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