[seqfan] Re: On equivalence of congruences

[Tomasz Ordowski]

Jean-Paul,

Thank you very much for this sketch of proof.
It convinces me, but others can ask for details.

Best regards,

Thomas

2018-07-29 21:11 GMT+02:00 jean-paul allouche <jean-paul.allouche at imj-prg.fr
>:

> Hi here is a brief sketch of a proof which (I hope) would work.
>
> First k^{p-1} == 1 if n not == 0 mod p and 0 otherwise.
> Hence the sum 1^{p-1} + ... + n^{p-1} is congruent to
> n - [n/p] mod p (here [x] stands for the integer part of x).
> Thus the sum is congruent to n mod p if and only if [n/p]
> is zero mod p. This is easily shown to be equivalent to
> there exists a \geq 0 and r \in [0,p-1] such that n = a p^2 + r.
>
> Now binomial(n+p,p) = (n+1)(n+2)...(n+p)/p!
> Let j be the integer in [1,p] such that n+j is divisible by p.
> Then binomial(n+p,p) = \prod_{k /= j} (n+k) . (n+j)/(p(p-1)!)
> == - \prod_{k /= j} (n+k) ((n+j)/p) mod p (use Wilson).
> I think that the product is also -1 mod p (kind of Wilson).
> So that binom(n+p,p) ==1 mod p if and only if (n+j)/p is
> congruent to 1 mod p, which is equivalent to saying that the
> remainder of the division of n by p^2 belongs to [0, p-1].
>
> Is this convincing?
> best
> jp
>
>
>
> Le 28/07/2018 à 08:17, Tomasz Ordowski a écrit :
>
>> Dear SeqFan,
>>
>> Conjecture:
>>
>> For a prime p;
>> 1^(p-1)+2^(p-1)+...+n^(p-1) == n (mod p)
>> if and only if
>> binomial(n+p,p) == 1 (mod p).
>>
>> I am asking for proof or counterexample.
>>
>> Cf. https://oeis.org/A133907
>>
>> ? a(n) is the smallest prime p such that
>> 1^(p-1)+2^(p-1)+...+n^(p-1) == n (mod p).
>>
>> Best regards,
>>
>> Thomas
>> _________
>> For n = p-1, binomial(2p-1,p) == 1 (mod p^3) with p>3.
>>
>> https://en.wikipedia.org/wiki/Wolstenholme%27s_theorem
>>
>> https://en.wikipedia.org/wiki/Agoh%E2%80%93Giuga_conjecture
>>
>>
>>
>>
>> <#m_8304939436458860045_m_1248764205802719950_m_253260159490
>> 715628_m_-6781035779772118739_m_4512355157857871716_DAB4FAD8
>> -2DD7-40BB-A1B8-4E2AA1F9FDF2>
>>
>> --
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>>
>
>
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