show that $(1^p)^k+(2^p)^k+\cdots+((p-1)^p)^k$ is divisible by $p$

Asked Dec 20 '18 at 16:48

Active Dec 20 '18 at 17:02

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This will be reduced to $1^k+2^k+3^k+\cdots+(p-1)^k$ by Fermat Theorem. I know that the sum of the reduced residue system of $n$ is divisible by $n$. What I need to show, though, is that $1^k+2^k+3^k+\cdots+(p-1)^k$ is a reduced residue system of $p$. How to do so?

Is there any other methods to prove so?

Added

I thought of pairing elements such as $a$ and $p-a \equiv -a$ to cancelled out. But this works for odd $k$ only.

edited Dec 20 '18 at 17:02

asked Dec 20 '18 at 16:48

Maged Saeed

1,140

2

This is not clear. Are you claiming that, say, $2^{p^k}\equiv 2^k\pmod p$? But this is false...$2^{3^2}=2^9\equiv 2\not \equiv 2^2\pmod 3$. – lulu Dec 20 '18 at 16:54
@DonThousand Unless I am misunderstanding something, $k≥0\implies a^{p^k}\equiv a \pmod p$...the $k$ has little meaning here. – lulu Dec 20 '18 at 16:57
by Fermat theorem, $a^p \equiv a \pmod p$ Hence, $a^{p^k} \equiv a^k \pmod p$. Is this not true? – Maged Saeed Dec 20 '18 at 16:59
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No. Keep in mind that $a^{p^k}$ denotes $a^{(p^k)}$ and not $\left( a^p\right)^k$ (which, after all, would just be $a^{pk}$). – lulu Dec 20 '18 at 17:01
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This problem is actually easier if you don't reduce using Fermat's theorem. You were on the right path with pairing, and for all primes greater than 2 your exponents will be odd. The case of 2 is a trivial case though. – mascoj Dec 20 '18 at 17:01
typo mistake though. – Maged Saeed Dec 20 '18 at 17:01
@lulu I have updated accordingly. – Maged Saeed Dec 20 '18 at 17:02
Oh nice @mascoj. I will get rid of even powers if I did not use $FT$. Thanks for appointing me to so. – Maged Saeed Dec 20 '18 at 17:04
However, can you show that $1^k,2^k,...,(p-1)^k$ is a reduced residue system? – Maged Saeed Dec 20 '18 at 17:05
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It is not a reduced residue system for $p=7$ and $k=3$. – lhf Dec 20 '18 at 17:07
@lhf But the sum is divisible by 7. – Maged Saeed Dec 20 '18 at 17:08
So, the appropriate question to ask is "show that $1^k+2^k+\cdots + (p-1)^k$ is divisible by $p$. – Maged Saeed Dec 20 '18 at 17:10
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Possible duplicate of $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$? – lulu Dec 20 '18 at 17:17
@lulu the question is closed. I was about to give the idea mentioned by mascoj there. Any way, Thanks,, – Maged Saeed Dec 20 '18 at 17:29
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Yes, it's closed (because the OP in that case posted no details)...but there are two excellent answers provided. – lulu Dec 20 '18 at 17:55
@lulu Yes yes I agree. But better to add more ideas since the prove is so short and elegant. – Maged Saeed Dec 20 '18 at 18:01
What's $p$ ?. Is it an integer ?. – Felix Marin Dec 20 '18 at 18:27
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@FelixMarin a prime. – Maged Saeed Dec 20 '18 at 19:43

show that $(1^p)^k+(2^p)^k+\cdots+((p-1)^p)^k$ is divisible by $p$

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