- 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

> Given a prime p its easy to see that

Warning: p must be an odd prime!

> it always divides 1+ 2 + .. + (p-1)

> But it in fact divides 1^k + 2^k + ... + (p-1)^k

It appears as a simple induction:

> 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?

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] - --- In primenumbers@yahoogroups.com, "S. R. Sudarshan Iyengar"

<sudarshanmysore@...> wrote:>

(p-1)

> 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 + .. +

>

is

> 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

> even?

This is not a property of the prime numbers but of the natural

>

> Regards,

> Sudarshan

>

numbers: put n instead of p, or, if you want, 2n-1 (odd numbers).

WDS