Prove that ## 1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p} ##.

[Math100]

Homework Statement

Employ Fermat's theorem to prove that, if $p$ is an odd prime, then
$1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p}$.
[Hint: Recall the identity $1+2+3+\dotsb +(p-1)=p(p-1)/2$.]

Relevant Equations

None.

Proof:

Suppose $p$ is an odd prime such that $p\geq 3$.
By Fermat's theorem, we have that $a^{p}\equiv a\pmod {p}$.
Then
\begin{align*}
&1^{p}\equiv 1\pmod {p}\\
&2^{p}\equiv 2\pmod {p}\\
&3^{p}\equiv 3\pmod {p}\\
&\vdots \\
&(p-1)^{p}\equiv (p-1)\pmod {p}.\\
\end{align*}
This means $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv [1+2+3+\dotsb +(p-1)]\pmod {p}$.
Since $1+2+3+\dotsb +n=\frac{n(n+1)}{2}$,
it follows that $1+2+3+\dotsb +(p-1)=\frac{p(p-1)}{2}$.
Observe that $p-1$ is even, so $p-1=2k$ for some $k\in\mathbb{N}$.
Now we have $1+2+3+\dotsb +(p-1)=pk$.
Thus $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv pk\equiv 0\pmod {p}$.
Therefore, if $p$ is an odd prime, then $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p}$.

 

[fresh_42]

Homework Statement:: Employ Fermat's theorem to prove that, if $p$ is an odd prime, then
$1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p}$.
[Hint: Recall the identity $1+2+3+\dotsb +(p-1)=p(p-1)/2$.]
Relevant Equations:: None.

Proof:

Suppose $p$ is an odd prime such that $p\geq 3$.
By Fermat's theorem, we have that $a^{p}\equiv a\pmod {p}$.
Then
\begin{align*}
&1^{p}\equiv 1\pmod {p}\\
&2^{p}\equiv 2\pmod {p}\\
&3^{p}\equiv 3\pmod {p}\\
&\vdots \\
&(p-1)^{p}\equiv (p-1)\pmod {p}.\\
\end{align*}
This means $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv [1+2+3+\dotsb +(p-1)]\pmod {p}$.
Since $1+2+3+\dotsb +n=\frac{n(n+1)}{2}$,
it follows that $1+2+3+\dotsb +(p-1)=\frac{p(p-1)}{2}$.

I think the following detour isn't necessary. As a sum of integers, $\frac{p(p-1)}{2}$ is automatically an integer, and $p$ obviously a factor. Hence $\frac{p(p-1)}{2} \equiv 0\pmod{p}.$

Observe that $p-1$ is even, so $p-1=2k$ for some $k\in\mathbb{N}$.
Now we have $1+2+3+\dotsb +(p-1)=pk$.
Thus $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv pk\equiv 0\pmod {p}$.
Therefore, if $p$ is an odd prime, then $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p}$.

 

[Math100]

I think the following detour isn't necessary. As a sum of integers, $\frac{p(p-1)}{2}$ is automatically an integer, and $p$ obviously a factor. Hence $\frac{p(p-1)}{2} \equiv 0\pmod{p}.$

I guess I wrote too much then.

 

[mathstudent1]

Homework Statement: Employ Fermat's theorem to prove that, if $p$ is an odd prime, then
$1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p}$.
[Hint: Recall the identity $1+2+3+\dotsb +(p-1)=p(p-1)/2$.]
Relevant Equations: None.

Proof:

Suppose $p$ is an odd prime such that $p\geq 3$.
By Fermat's theorem, we have that $a^{p}\equiv a\pmod {p}$.
Then
\begin{align*}
&1^{p}\equiv 1\pmod {p}\\
&2^{p}\equiv 2\pmod {p}\\
&3^{p}\equiv 3\pmod {p}\\
&\vdots \\
&(p-1)^{p}\equiv (p-1)\pmod {p}.\\
\end{align*}
This means $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv [1+2+3+\dotsb +(p-1)]\pmod {p}$.
Since $1+2+3+\dotsb +n=\frac{n(n+1)}{2}$,
it follows that $1+2+3+\dotsb +(p-1)=\frac{p(p-1)}{2}$.
Observe that $p-1$ is even, so $p-1=2k$ for some $k\in\mathbb{N}$.
Now we have $1+2+3+\dotsb +(p-1)=pk$.
Thus $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv pk\equiv 0\pmod {p}$.
Therefore, if $p$ is an odd prime, then $1^{p}+2^{p}+3^{p}+\dotsb +(p-1)^{p}\equiv 0\pmod {p}$.

what about if p=2? because i have homework and in the question, they said for any prime p

 

[Steve4Physics]

what about if p=2? because i have homework and in the question, they said for any prime p

$1^p + 2^p +3^p ~+ ... +~(p-1)^p \equiv 0~(\text {mod } p)$ won't work if $p=2$. Can you see why? What is the exact wording of the question?

(BTW, the original thread is over 1 year old.)

 

[mathstudent1]

$1^p + 2^p +3^p ~+ ... +~(p-1)^p \equiv 0~(\text {mod } p)$ won't work if $p=2$. Can you see why? What is the exact wording of the question?

(BTW, the original thread is over 1 year old.)

because 1 is not congruent to 0 mod 2. so, this prove holds only for odd primes? the reason that I ask is because my teacher wrote for any prime proof that

$1^p + 2^p +3^p ~+ ... +~(p-1)^p \equiv 0~(\text {mod } p)$

 

[Steve4Physics]

because 1 is not congruent to 0 mod 2.

Yes.

so, this prove holds only for odd primes?

Yes. (Of course we note that all primes are odd except for 2.)

(However, beware of terminology. The term 'even primes' is sometimes, very confusingly IMO, used to mean the sequence 2, 4, 6, 10, 14, 22, 26, 34, 38 .... E.g. see here.)

the reason that I ask is because my teacher wrote for any prime proof

Then I'd say it's a mistake/oversight.