Sorry, but posted at least two times before.

I would like to see the links, because i suspected it was well known

Hint: group $1/a^p+1/(p-a)^p$, use the binomial theorem and rule out the $p^2$ factor, we are left with a sum mod $p$, and we must show it is $0$ mod $p$, this sum is related to the sum of squares mod $p$ which is $0$.

Well, sometimes I think it's easier to put up a solution than searching... I suspected that it was well known too. But I solved it in a slightly different way. I used the extended binomial expansion, that is, \[ (a + b)^\alpha = \sum_{k=0}^\infty{\alpha \choose k}a^{\alpha-k}b^k \] for $\alpha$ real. We define ${\alpha \choose k}$ for $\alpha$ real and $k$ nonnegative integer as ${\alpha \choose 0} = 1$ and \[ {\alpha\choose k} = \dfrac{\alpha(\alpha-1)(\alpha-2)\ldots (\alpha - k + 1)}{k!} \] for $k > 0$. So \[ a^{-p} + (p-a)^{-p} = a^{-p} + \sum_{k=0}^\infty {-p\choose k}(-a)^{-p-k}p^k \equiv a^{-p} - a^{-p} + {-p\choose 1}a^{-p-1}p - {-p\choose 2}a^{-p-2}p^2 \pmod{p^3} \] It's not hard to see that $p$ divides ${-p\choose 2}$, so \[ a^{-p} + (p-a)^{-p} \equiv -a^{-p-1}p^2\pmod{p^3} \] OK, it looks like "cheating" but it's not. You can check by hand that, indeed, \[ (p-a)^p \cdot (-a^{-p}-a^{-p-1}p^2) \equiv 1 \pmod{p^3} \] In fact, using the "regular" binomial expansion, \[ (p-a)^p \equiv -a^p + {p\choose 1}a^{p-1}p \pmod p \] and \[ (p-a)^p \cdot (-a^{-p}-a^{-p-1}p^2) \equiv (-a^p + a^{p-1}p^2)\cdot (-a^{-p}-a^{-p-1}p^2) \equiv 1 + a^{-1}p^2 -a^{-1}p^2\equiv 1 \pmod{p^3} \] Thus we only need to prove that \[ \sum_{k=1}^{p-1}a^{-p-1}\equiv 0\pmod p \] or, recalling that $a^{-p-1} \equiv (a^{-1})^{p+1} \equiv (a^{-1})^2\pmod p$, \[ \sum_{k=1}^{p-1}(a^{-1})^2\equiv 0\pmod p \] This is a standard problem in number theory. Let $S = \sum_{k=1}^{p-1}(a^{-1})^2\bmod p$. Thus, since $\mathbb{Z}/p\mathbb{Z}^* = \{1,2,\ldots, p-1\}$ is a reduced residue system, $2^{-2}\mathbb{Z}/p\mathbb{Z}^* = \mathbb{Z}/p\mathbb{Z}^*$ and \[ 2^{-2}S \equiv S\pmod p\iff S \equiv 0 \pmod p, \] since $p \neq 3$ and we are done.

cyshine wrote: Let $p > 3$ be a prime. Prove that if \[ \sum_{i=1}^{p-1}{1\over i^p} = {n\over m}, \] with $\gdc(n,m) = 1$, then $p^3$ divides $n$. Raise Wolstenholme's congruence (mod $p^2$) to the $p$'th power.

why can you raise it? i mean, i know exponentation rule...but this?

Other solution: Let $P(x)=(x-1^p)(x-2^p)...(x-(p-1)^p)=x^{p-1}+x^{p-2}A_1...+xA_{p-2}+\prod_{i=1}^{p-1}i^p$ By a standard result and since $a^p=a$ modp, we have that $A_1,A_2...A_{p-2}$ are multiples of $p$. Now put take $P(p^2)-\prod_{i=1}^{p-1}i^p$, we will show that $p^5$ divides it. Iff $p^5$ divides it, then $p^5$ divides the sum $p^{2(p-1)}+p^{2(p-2)}A_1...+p^2A_{p-2}$ which implies that $p^3$ divides $A_{p-2}$ (this implies of course what we wanted) Now we have that $(p^2-a^p)(p^2-(p-a)^p)=p^4-p^2(a^p+(p-a)^p)+a^p(p-a)^p$ But $p^5$ divides $p^4-p^2(a^p+(p-a)^p)$ by a simple binomial expansion of $(p-a)^p$. So we get that $(p^2-a^p)(p^2-(p-a)^p)=p^4-p^2(a^p+(p-a)^p)+a^p(p-a)^p=a^p(p-a)^p$(mod$p^5$) So we get that $P(p^2)-\prod_{i=1}^{p-1}i^p$ is a multiple of $p^5$ like we wanted.

The previus post inmediately generalizes to further results.

Pascual2005 wrote: why can you raise it? i mean, i know exponentation rule...but this? I think you are right, that it is not so simple. I had in mind that raising to the power $p$ tends to improve congruences mod $p^k$ to congruences modulo $p^{k+1}$, but did not check carefully the details. If it does work I will repost with the details.

This is equivalent to wishing to prove that \[2\sum_{i=1}^{p-1}\frac{1}{i^p}\equiv 0\pmod{p^3}\] treating each of the numbers as inverses and using $\gcd(p,2)=1$. We now notice $\frac{1}{j^p}+\frac{1}{(p-j)^p}=\frac{(p-j)^p+j^p}{j^p(p-j)^p}$. Therefore \[2\sum_{i=1}^{p-1}\frac{1}{i^p}=\sum_{i=1}^{p-1}\frac{(p-i)^p+i^p}{i^p(p-i)^p}\] By lifting the exponent $v_p\left( (p-i)^p+i^p\right)=v_p\left( p-i+i\right)+v_p(p)=2$. Therefore we can factor a $p^2$ out of the numerator to give us \begin{align*} \sum_{i=1}^{p-1}\frac{(p-i)^p+i^p}{i^p(p-i)^p}=p^2\sum_{i=1}^{p-1}\frac{\frac{(p-i)^p+i^p}{p^2}}{i^p(p-i)^p} \equiv 0\pmod{p^3}\\\iff \sum_{i=1}^{p-1}\frac{\frac{(p-i)^p+i^p}{p^2}}{i^p(p-i)^p}\equiv 0\pmod{p}\end{align*} Notice that $i\equiv -\left(p-i\right)\pmod{p}$ so hence we now need \[\sum_{i=1}^{p-1}\frac{\frac{(p-i)^p+i^p}{p^2}}{i^{2p}}\equiv 0\pmod{p}\] since we can take the negative out of the denominator. We consider \[\sum_{i=1}^{p-1}\frac{\frac{(\frac{p+1}{2})^p+(\frac{p-1}{2})^p}{p^2}}{i^{2p}}+\sum_{i=1}^{p-1}\frac{\frac{i^p+(p-i)^p-(\frac{(p+1)}{2})^p-(\frac{(p-1)}{2})^p}{p^2}}{i^{2p}}\] We desire to show each part is divisible by $p$. For the first sum notice that we want $\sum_{i=1}^{p-1}\frac{1}{i^{2p}}\equiv \sum_{i=1}^{p-1}i^{2p}\pmod{p}$ since each element from $1$ to $p-1$ has a distinct inverse including $1^{-1}\equiv 1\pmod{p}$ and $(p-1)^{-1}\equiv (p-1)\pmod{p}$. By Fermat's Little Theorem we have \[\sum_{i=1}^{p-1}i^{2p}\equiv \sum_{i=1}^{p-1}i^2\pmod{p}\equiv \frac{(p-1)(p)(2p-1)}{6}\equiv 0\pmod{p}\] since $p\neq 2,3$. Now we desire to prove that $\frac{i^p+(p-i)^p-\left(\frac{(p+1)}{2}\right)^p-\left(\frac{p-1}{2}\right)^p}{p^2}\equiv 0\pmod{p}$. Since $\gcd(p,2)=1$ we want \begin{align*} 2^pi^p+2^p(p-i)^p-(p+1)^p-(p-1)^p\equiv 0\pmod{p^3} \\ 2^p\left[ i^p+\binom{p}{1}p(-i)^{p-1}+(-i)^{p}\right]-\left[p\binom{p}{1}+1+p\binom{p}{1}(-1)^{p-1}+(-1)^{p}\right]\equiv 0\pmod{p^3} \\ 2^p\left[p^2\right]-2p^2\equiv0\pmod{p^3} \end{align*} The second step follows from the fact that $p^2\binom{p}{2}\equiv 0\pmod{p^3}$ and every $i\ge 3$ we have $p^i\binom{p}{i}\equiv 0\pmod{p^3}$, and the third step follows from $p$ being odd. Now, $p^2\left(2^p-2\right)\equiv 0\pmod{p^3}$ is true by Fermat's Little Theorem therefore we are done $\blacksquare$.

Hopefully this isn't a fakesolve Someone please check. Solution. It's equivalent to show, similarily as above, that $$2\sum_{i=1}^{p-1} \frac{1}{i^p} \equiv 0 \pmod{p^3}$$which is, by pairing \begin{align*} 2\sum_{i=1}^{p-1} \frac{1}{i^p} =\sum_{i=1}^{p-1} \frac{i^p+(p-i)^p}{i^p(p-i)^p} &= \sum_{i=1}^{p-1} \frac{\left[\dbinom{p}{0}p^p - \dbinom{p}{1}p^{p-1}i + \ldots - \dbinom{p}{p-2}p^2i^{p-2}+ \dbinom{p}{p-1}pi^{p-1}\right]}{i^p(p-i)^p} \\ &\equiv \sum_{i=1}^{p-1}\frac{\dbinom{p}{p-1}pi^{p-1}}{i^p(p-i)^p} \pmod{p^3} \\ &= \sum_{i=1}^{p-1} \frac{p^2i^{p-1}}{i^p(p-i)^p}=p^2 \sum_{i=1}^{p-1}\frac{1}{i(p-i)^p} \\ \end{align*}Therefore, it suffices to show \begin{align*} \sum_{i=1}^{p-1} \frac{1}{i(p-i)^p} \equiv 0 \pmod{p} &\iff \sum_{i=1}^{p-1} \frac{1}{-i^{p+1}} \equiv 0 \pmod{p} \\ &\iff \sum_{i=1}^{p-1} -\frac{1}{i^2} \equiv 0 \pmod{p} \iff \sum_{i=1}^{p-1} -i^2 \equiv 0 \pmod{p} \\ \end{align*}but last statement clearly holds since $$-\frac{(p-1)(p)(2p-1)}{6} \equiv 0 \pmod{p}$$which finishes the problem. $\blacksquare$

"Wolstenholme take me home to the place I belong West Virginia mountain mamma take me home Wolstenholme." Shout out to John Denver A huge motivation for solving this problem is Wolstenholme’s Theorem, which is stated as follows Theorem (Wolstenholme): Let $p>3$ be prime. Then if $$\sum_{i=1}^{p-1}\frac{1}{i}=\frac{m}{n}$$for $\gcd(m,n)=1$ we have $p^2~|~m$. An alternative statement is simply that $$\sum_{i=1}^{p-1}\frac{1}{i}\equiv 0\pmod{p^2}.$$A proof of this theorem will provide some motivation on how we should approach this problem. We have \begin{align*} 2\sum_{i=1}^{p-1}\frac{1}{i}&=\sum_{i=1}^{p-1}\left(\frac{1}{i}+\frac{1}{p-i}\right)\\ &=\sum_{i=1}^{p-1}\frac{p}{i(p-i)}\\ &\equiv p\sum_{i=1}^{p-1}-\frac{1}{i^2}\\ &\equiv-p\sum_{i=1}^{p-1}i^2\\ &=-\frac{p^2(p-1)(2p-1)}{6}\\ &\equiv 0\pmod{p^2} \end{align*}which completes the proof. So how does this motivate our solution? We notice that the problem statement is equivalent to proving that $$2\sum_{i=1 }^{p-1}{1\over i^p}\equiv 0\pmod{p^3}$$so we are basically proving Wolstenholme, but with a couple of more special conditions to think about. Using an identical argument we see that \begin{align*} 2\sum_{i=1 }^{p-1}{1\over i^p}=\sum_{i=1}^{p-1}\frac{(p-i)^p+i^p}{i^p(p-i)^p}\\ \end{align*}and using the fact that $$(p-i)^p=\sum_{k=0}^{p}\binom{p}{k}p^{p-i}(-i)^k$$our sum simplifies to \begin{align*} \sum_{i=1}^{p-1} \frac{i^p+(p-i)^p}{i^p(p-i)^p} &\equiv \sum_{i=1}^{p-1}\frac{\dbinom{p}{p-1}pi^{p-1}}{i^p(p-i)^p}\\ &\equiv\sum_{i=1}^{p-1} \frac{p^2i^{p-1}}{i^p(p-i)^p}\\ &\equiv p^2 \sum_{i=1}^{p-1}\frac{1}{i(p-i)^p}\pmod{p^3}. \end{align*}Similar to how we did it in proving Wolstenholme, we have to prove that $$\sum_{i=1}^{p-1}\frac{1}{i(p-i)^p}\equiv 0\pmod{p}$$since that is easier to prove and generalizes to $p^3$. Using the fact that $p-i\equiv -i\pmod{p}$ again and some smooth slight of hand we get \begin{align*} \sum_{i=1}^{p-1}\frac{1}{i(p-i)^p}&\equiv \sum_{i=1}^{p-1}-\frac{1}{i^{p+1}}\pmod{p}\\ &\equiv \sum_{i=1}^{p-1}-\frac{1}{i^{2}}\pmod{p}\\ &\equiv -\sum_{i=1}^{p-1}i^2\\ &\equiv 0\pmod{p}\\ \end{align*}which finishes just like Wolstenholme.

It suffices to show $\sum_{i=1}^{p-1}\frac{1}{i^p}\equiv0\pmod{p^3}.$ Notice \begin{align*}&0\equiv\sum_{i=1}^{p-1}\frac{1}{i^p}\equiv\sum_{i=1}^{\frac{p-1}{2}}\left(\frac{1}{i^p}+\frac{1}{(p-i)^p}\right)\equiv \sum_{i=1}^{\frac{p-1}{2}}\frac{i^p+(p-i)^p}{i^p(p-i)^p}\pmod{p^3}\\&\iff0\equiv\sum_{i=1}^{\frac{p-1}{2}}\frac{i^p+p^p+\dots+\binom{p}{2}p^2(-i)^{p-2}+\binom{p}{1}p(-i)^{p-1}+(-i)^p}{i^p(p-i)^p}\equiv\sum_{i=1}^{\frac{p-1}{2}}\frac{p^2i^{p-1}}{i^p(p-i)^p}\pmod{p^3}\\&\iff0\equiv\sum_{i=1}^{\frac{p-1}{2}}\frac{p^2}{i(p-i)^p}\pmod{p^3}\iff0\equiv\sum_{i=1}^{\frac{p-1}{2}}\frac{1}{i(p-i)^p}\pmod{p}\\&\iff0\equiv2\sum_{i=1}^{\frac{p-1}{2}}\frac{1}{-i^{p+1}}\equiv\sum_{i=1}^{p-1}\frac{1}{i^2}\pmod{p}\\&\iff0\equiv\sum_{i=1}^{p-1}i^2\equiv\frac{p(p-1)(2p-1)}{6}\pmod{p},\end{align*}which is true. $\square$

Mogmog8 wrote: $$2\sum_{i=1}^{\frac{p-1}{2}}\frac{1}{-i^{p+1}}\equiv\sum_{i=1}^{p-1}\frac{1}{i^2}\pmod{p}$$ How is this true ?

Let $p > 3$ be a prime. Prove that if $$\sum_{i=1 }^{p-1}{1\over i^p} = {n\over m}, $$with $ gcd\,\, (n,m) = 1$, then $p^3$ divides $n$.

parmenides51 wrote: Let $p > 3$ be a prime. Prove that if $$\sum_{i=1 }^{p-1}{1\over i^p} = {n\over m}, $$with $ gdc\,\, (n,m) = 1$, then $p^3$ divides $n$.

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Hi! I will like to add a minor correction. It's actually $\gcd (n,m)$ and not $gdc(n,m)$, the original post has a typing mistake there.

Let $S=\frac{n}{m}.$ We want to show that $2S = 0$ modulo $p^3$. To do this notice that $$2S=\sum_{i=1}^{p-1}\frac{1}{i^p}+\frac{1}{(p-i)^p}=\sum_{i=1}^{p-1}\frac{i^p+(p-i)^p}{i^p(p-i)^p}.$$See that, by Newton's Binomial, that $$i^p+(p-i)^p\equiv -\binom{p}{2}p^2i^{p-2}+\binom{p}{1}pi^{p-1} \mod p^3.$$But as obviously $p\mid \binom{p}{2}$, we obtain that the expresion is just $p^2i^{p-1}$ modulo $p^3$. So now, showing that our sum is $0$ modulo $p^3$ is equivalent to showing that the following is $0$ modulo $p$ $$\sum_{i=1}^{p-1} \frac{i^{p-1}}{i^p(p-i)^p}\equiv \sum_{i=1}^{p-1} \frac{1}{-i^2}\equiv-\sum_{j=1}^{p-1}j^2\equiv -p \frac{(p+1)(2p+1)}{6},$$which is $0$ modulo $p$ for $p>3$, which ends the problem.

Because $p > 3$, $p$ is an odd prime and thus $p - 1$ is even. This allows us to pair up terms like so: \[\sum_{i = 1}^{\frac{p - 1}{2}} \frac{1}{i^p} + \frac{1}{(p - i)^p}\]\[\sum_{i = 1}^{\frac{p - 1}{2}} \frac{i^p + (p-i)^p}{(i(p-i))^p}\] From the binomial theorem, the numerator is equal to \[N = i^p + p^p - {p \choose 1}ip^{p - 1} + \dots + {p \choose p - 1}i^{p-1}p - i^p\]\[N = p^p - {p \choose 1}ip^{p - 1} + \dots + {p \choose p - 1}i^{p-1}p\] Note that since $p$ is prime, $p \mid {p \choose k}$ for all $1 \leq k \leq p - 1$; we thus conclude that $p^2 \mid N$. We now must show that \[p \mid \sum_{i = 1}^{\frac{p - 1}{2}} \frac{1}{(i(p-i))^p}\] Rewrite the sum as \[-\sum_{i = 1}^{p - 1} \frac{1}{i^{2p}}\] modulo $p$. Because each number has a unique inverse modulo $p$, we can determine this sum is congruent to \[-\sum_{i = 1}^{p - 1} i^{2p}\] modulo $p$. Using FLT, this sum reduces to \[-\sum_{i = 1}^{p - 1} i^2 = -\frac{(p - 1)p(2p - 1)}{6}\] Because $p > 3$, we can thus conclude that \[p \mid -\frac{(p - 1)p(2p - 1)}{6}\] as desired.