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## A nice problem...

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• Here goes a nice problem... IT would be nice if someone can give me links/references where I can find more details about this problem. Given a prime P its easy
Message 1 of 3 , Dec 27, 2007
Here goes a nice problem... IT would be nice if someone can give me
links/references where I can find more details about this problem.

Given a prime P its easy to see that it always divides 1+ 2 + .. + (p-1)

But it in fact divides 1^k + 2^k + ... + (p-1)^k for all k except 0
and p-1.

Its obvious if k is odd. But is there any way of proving it when k is
even?

Regards,
Sudarshan
• Happy new year to all ... Warning: p must be an odd prime! ... It appears as a simple induction: Let S(n) denote 1^n+ 2^n + ... + (p-1)^n Recalling the
Message 2 of 3 , Dec 27, 2007
Happy new year to all

> Given a prime p its easy to see that
> it always divides 1+ 2 + .. + (p-1)
Warning: p must be an odd prime!

> But it in fact divides 1^k + 2^k + ... + (p-1)^k
> for all k except 0 and p-1.

> Its obvious if k is odd.
> But is there any way of proving it when k is even?

It appears as a simple induction:

Let S(n) denote 1^n+ 2^n + ... + (p-1)^n

Recalling the binomial expansion
(a+1)^n = a^n + (n|1)a^(n-1)+ ... + (n|n)
where (n|m) denotes the binomial coefficients n!/m!(n-m)!

Summing it up for a=1,...,p-1 and after telescoping ^n terms from both sides
of the equality we obtain a recursive relation for S(n-1).

Replacing n with k+1 we have a recursive relation for S(k) as follows:

p^(k+1) = 1 + (k+1|1)S(k) + (k+1|2)S(k-1) + ... + (k+1|k)S(1) + (p-1)

Now the induction works independent of parity of k. Please fill the small
details.

Regards
Payam

[Non-text portions of this message have been removed]
• ... (p-1) ... is ... This is not a property of the prime numbers but of the natural numbers: put n instead of p, or, if you want, 2n-1 (odd numbers). WDS
Message 3 of 3 , Dec 31, 2007
--- In primenumbers@yahoogroups.com, "S. R. Sudarshan Iyengar"
<sudarshanmysore@...> wrote:
>
> Here goes a nice problem... IT would be nice if someone can give me
> links/references where I can find more details about this problem.
>
> Given a prime P its easy to see that it always divides 1+ 2 + .. +
(p-1)
>
> But it in fact divides 1^k + 2^k + ... + (p-1)^k for all k except 0
> and p-1.
>
> Its obvious if k is odd. But is there any way of proving it when k
is
> even?
>
> Regards,
> Sudarshan
>

This is not a property of the prime numbers but of the natural
numbers: put n instead of p, or, if you want, 2n-1 (odd numbers).

WDS
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