It suffices to show $\dbinom{p}{p-1}x^{p-1} + \dbinom{p}{p-2}x^{p-2} + ... + \dbinom{p}{1}x \equiv 0 \pmod{p^2}$ by multiplying the conclusion we want by $p$. So we want then: \[\sum_{i=0}^{p} \dbinom{p}{i} x^i \equiv x^p + 1 \pmod{p^2}\] \[ (1+x)^p \equiv x^p + 1 \pmod{p^2}\] Now it is a well-known problem that when $p|(a^2 + ab + b^2)$ then $p^3|((a+b)^p - a^p - b^p)$ for $p \ge 5$. Then since that $p|(x^2 + x + 1)$, we are done.

dinoboy, could you please explain your first step? I did not see why one implies the other...

I'm using the (highly) useful identity $\frac{(-1)^{k-1}p}{k} \equiv \dbinom{p}{k} \pmod{p^2}$

Ok, now I've got it! Really nice identity Indeed, it's a know example of the Wilson's Theorem, now I realize there are many books with this relation. Also, nice proof for the problem

Just for the sake of it, another finish, even though really the first step is the key. LEMMA. For $k=1,2,\dots,p-1$ we have \[ \frac{1}{k} \equiv (-1)^{k-1} \frac{1}{p} \binom{p}{k} \pmod p. \]Proof. Not too difficult. $\square$ Note that it suffices to just take the second factor and let the fraction be inverses, using the lemma we have \[ x - \frac {x^2}{2} + \frac {x^3}{3} - \cdots - \frac {x^{p - 1}}{p - 1} \equiv \frac{1}{p} \sum_{i=1}^{p-1} x^i \binom{p}{i} = \frac{1}{p} \left( (x+1)^p-x^p-1 \right). \]Note that the requirement $p \mid x^3-1$ but $p \nmid x-1$ gives us $p \mid x^2+x+1$, so $x+1 \equiv -x^2 \pmod p$ which lets us transform the above into \[ \frac{1}{p} \left( (x+1)^p-x^p-1 \right) = - \frac{1}{p} \left( x^{2p}+x^p+1 \right) = - \frac{1}{p} \cdot \frac{x^{3p}-1}{x-1}. \]As $p \nmid x-1$, it now suffices to prove $p^2 \mid x^{3p}-1$. For that use \[ x^{3p-1}-1 = \left(x^3-1 \right) \left(x^{3p-3}+x^{3p-6}+\dots+1 \right). \]The first factor is divisible by $p$, so if we can show that the second factor is divisible by $p$ as well, then we are done. But using $x^3 \equiv 1 \pmod p$ we get \[ x^{3p-3}+x^{3p-6}+\dots+1 \equiv \underbrace{1+1+\dots+1}_{p \text{ times}} = p \equiv 0 \pmod p. \]Done. $\square$

Nice and easy! Here's my solution: Note that, for any $1 \leq i \leq p-1$, we have $$\frac{1}{i}=\frac{(p-i)!(i-1)!}{i!(p-i)!}=\frac{(i-1)!(p-(p-1))(p-(p-2)) \dots (p-i)}{i!(p-i)!} \equiv \frac{(-1)^{p-i} (p-1)!}{i!(p-i)!}=\frac{(-1)^{i-1}}{p} \binom{p}{i} \pmod{p}$$Using Wilson's Theorem, and the above fact, it suffices to prove $$0 \equiv (p-1)! \sum_{i=1}^{p-1} \frac{(-1)^ix^i}{i} \equiv \sum_{i=1}^{p-1} (-1)^{i+1}x^i \times \frac{(-1)^{i-1}}{p} \binom{p}{i} \Leftrightarrow \sum_{i=1}^{p-1} \binom{p}{i} x^i \equiv 0 \pmod{p^2}$$Since $x^3 \equiv 1 \pmod{p}$ (which also gives $3 \mid p-1$, since $3$ is the order of $x$ modulo $p$), so we get that $$\frac{1}{p}\sum_{i=1}^{p-1} \binom{p}{i} x^i \equiv \frac{1}{p} \left(\sum_{j=1}^{\frac{p-1}{3}} \binom{p}{3j}+x\sum_{j=1}^{\frac{p-1}{3}} \binom{p}{3j-2}+x^2\sum_{j=1}^{\frac{p-1}{3}} \binom{p}{3j-1} \right) \pmod{p}$$Let $\omega$ denote a complex cube root of unity. Then one easily gets (using the expansion of $(1+x)^p$, and from $1+\omega+\omega^2=0$, as well as by $p \equiv 1 \pmod{3}$) that $$a=\left(\sum_{j=0}^{\frac{p-1}{3}} \binom{p}{3j} \right)-\binom{p}{0}=\frac{2^p+(1+\omega)^p+(1+\omega^2)^p}{3}-1=\frac{2^p-\omega^{2p}-\omega^p}{3}-1=\frac{2^p-\omega^2-\omega-3}{3}=\frac{2}{3}(2^{p-1}-1)$$$$b=\left(\sum_{j=1}^{\frac{p+2}{3}} \binom{p}{3j-2} \right)-\binom{p}{p}=\frac{2^p+\omega^2(1+\omega)^p+\omega(1+\omega^2)^p}{3}-1=\frac{2^p-\omega^{2p+2}-\omega^{p+1}}{3}-1=\frac{2^p-\omega-\omega^2-3}{3}=\frac{2}{3}(2^{p-1}-1)$$$$c=\sum_{j=1}^{\frac{p-1}{3}} \binom{p}{3j-1}=\frac{2^p+\omega(1+\omega)^p+\omega^2(1+\omega^2)^p}{3}=\frac{2^p-\omega^{2p+1}-\omega^{p+2}}{3}-1=\frac{2^p-1-1}{3}=\frac{2}{3}(2^{p-1}-1)$$Thus, we find that $$a+bx+cx^2 =\frac{2}{3}(2^{p-1}-1)(1+x+x^2) \equiv 0 \pmod{p^2}$$where the last congruence follows by FLT, and the fact that $\text{ord}_p(x)=3$ implies $p \mid 1+x+x^2$. Hence, done. $\blacksquare$

$p \geq 7$

I don't think I used the fact that $p\nmid x-1$. Using $\frac{1}{k}\equiv (-1)^{k-1}\frac{1}{p}\binom{p}{k}\pmod p$, we get that $$(p-1)!\sum_{k=1}^{p-1}(-1)^{k-1}\frac{x^k}{k}\equiv (p-1)!\sum_{k=1}^{p-1}\frac{1}{p}\binom{p}{k}x^k\equiv \sum_{k=1}^{p-1}\binom{p-1}{k}x^k\equiv (x+1)^{p-1}-1\pmod p.$$Now since $x^3\equiv 1\pmod p$, so $x\not\equiv -1\pmod p$, since $p>2$. So $\gcd(x+1,p)=1$. Then by FLT, $(x+1)^{p-1}-1\equiv 0\pmod p$, and we are done.

redacted

Sorry for reviving. Does anyone know if $p^3|((a+b)^p - a^p - b^p)$ is satisfied, then it implies that $p|(a^2 + ab + b^2)$ is satisfied also? Thank you!

It is not necessarily true. Take $a\equiv -1\pmod{p^3}$ and $b\equiv 1\pmod{p^3}$. Then $p^3$ clearly divides $(a+b)^p-a^p-b^p$ but $a^2+ab+b^2\equiv1\pmod{p}$. However, what is the proof for this: Quote: Now it is a well-known problem that when $p|(a^2 + ab + b^2)$ then $p^3|((a+b)^p - a^p - b^p)$ for $p \ge 5$ I suspect it has to do with Hensel's lemma EDIT: @dinoboy only used the fact fact that if $p|(a^2 + ab + b^2)$ then $p^2|((a+b)^p - a^p - b^p)$, which can be proven rather easy with some tricks involving Wilson's theorem, but for $p^3$ it is quite hard. As an observation, to prove that if $p|(a^2 + ab + b^2)$ then $p^3|((a+b)^p - a^p - b^p)$, it is enough to prove that if $p|k^2+k+1$, then $p^3|(k+1)^p-k^p-1$

Vlad-3-14 wrote: However, what is the proof for this: Quote: Now it is a well-known problem that when $p|(a^2 + ab + b^2)$ then $p^3|((a+b)^p - a^p - b^p)$ for $p \ge 5$ You can find some proofs here and here.

Thank you @above.

We work in $\mathbb{F}_p$ so as to discard the $(p-1)!$ term. Remark that $p\ne 3$. Additionally remark that the condition implies there is an element of order $3$ in $\mathbb{Z}_p^{\times}$, $x$, so $6\mid p-1$. By the harmonic mod $p$ trick, the sum is \[\tfrac 1p \sum_{k=1}^{p-1}x^k\binom{p}{k} = \tfrac 1p((1+x)^p-1-x^p).\]To show $(1+x)^p-1-x^p$ is divisible by $p^2$, we show it is divisible by $(x^2+x+1)^2$. Let $\omega=e^{2i\pi/3}$. Compute that \[(1+\omega)^p-1-\omega^p=(1+\omega)-1-\omega=0,\]so $x-\omega\mid (1+x)^p-1-x^p$. Compute that \[p(1+\omega)^{p-1}-p\omega^{p-1}=p-p=0,\]so $(x-\omega)^2\mid (1+x)^p-1-x^p$. This implies the desired.

The sum computes in $[\mathbb{F}_p]$ to $\sum_{i=1}^{p-1} \frac{(-1)^{i+1}x^i}{i} \equiv \sum_{i=0}^{p} \dbinom{p}{i} x^i \equiv {(x+1)^p-x^p-1 \over p} \pmod p$ Now we will show that $p^2|(x+1)^p-x^p-1$ Denote $X$ to be the inverse of $x$ modulo $p$. Since $p|X^{3p}-1$ and so $p^2|(X^p-1)(X^{2p}+X^p+1)$ or $p^2|X^{2p}+X^p+1$ Since $X+1 \equiv -X^2 \pmod p$,we must have $(X+1)^p \equiv X^p+1 \pmod p$, implying the result.

Quite a nice problem you know!Though a bit wrong(p=3 doesn't work) Firstly (and obviously) note that by Wilson's, $p|(p-1)!+1$ So we'll show that p divides $[ x - \frac {x^2}{2} + \frac {x^3}{3} - \cdots - \frac {x^{p - 1}}{p - 1}$ Using $\frac{1}{k}\equiv (-1)^{k-1}\frac{1}{p}\binom{p}{k}\pmod p$ We'll show that $p$ divides $\frac{1}{p} \sum_{i=1}^{p-1} x^i \binom{p}{i}$ So we'll show that $p^2$ divides $\left( (x+1)^p-x^p-1 \right)$ (by binomial theorem) Since $x+1=pk-x^2$ for some k ,expanding we just need to show that $p|x^{2p} + x^p +1$ We know that $p \mid x^3 - 1$ and by the well known lemma $a\equiv b\pmod{m^n}$ implies $a^m\equiv b^m\pmod{m^{n+1}}$ we get that $p^2|x^{3p} -1$ Now all we need to do is to show that ($x^p -1$,p)=1 .Now assume to the contrary that this is true and p>3 .Notice that either p-1 or p+1 is a multiple of 3.Note that $p \mid x^3 - 1$ Now just use powers over congruences and subtract the 2 results.We get a contradiction.Hence $p|x^{2p} + x^p +1$ Thus we have proved the result

I'm quite surprised to see that all the solutions so far depend on the "explicit" evaluation of the sum by the "harmonic mod p trick". There is an alternative "functional equation" route which I find very natural: Let $f(z)=z+\frac{z^2}{2}+\dots+\frac{z^{p-1}}{p-1}$ where we think of $f$ as a map from $\mathbb{F}_p \to \mathbb{F}_p$. We want to prove $f(-x)=0$ whenever $x$ has order $3$. As explained here (this is the 2011-N7 thread) one readily checks that we have the "functional equations" \[f(z+1)=f(-z)\]and \[f(z^2)=(1+z)f(z)+(1-z)f(-z).\]From there one can finish quickly: Note that $x^2+x+1=0$ and hence $f(-x^2)=f(x+1)=f(-x)$ from the first functional equation. Applying the second one first with $z=x^2$ and then again with $z=x$ we get \[f(x)=(1+x^2)f(x^2)+(1-x^2)f(-x^2)=-xf(x^2)+(1-x^2)f(-x)=-x((1+x)f(x)+(1-x)f(-x))+(1-x^2)f(-x)=f(x)+(1-x)f(-x)\]and hence $(1-x)f(-x)=0$ and hence $f(-x)=0$ as desired.

We use the following "trick": for $k \not \equiv 0 \pmod{p}$, we have $$\frac{1}{k} \equiv (-1)^{p-1}\cdot \frac{1}{p}\binom{p}{k} \pmod{p}.$$Applying this, it suffices to show that \begin{align*} \sum_{k=1}^{p-1} \binom{p}{k}x^k &\equiv 0 \pmod{p^2}\\ \sum_{k=0}^p \binom{p}{k}x^k &\equiv x^p+1 \pmod{p^2}\\ (x+1)^p \equiv x^p+1 \pmod{p^2}. \end{align*}From $p \mid x^3-1$ but $p \nmid x-1$, we get $p \mid x^2+x+1$, so by Exponent Lifting $$\nu_p((x+1)^p+x^{2p})=1+\nu_p(x^2+x+1) \geq 2 \implies (x+1)^p \equiv -x^{2p} \pmod{p^2},$$so it suffices to prove that $p^2 \mid x^{2p}+x^p+1=\tfrac{x^{3p}-1}{x^p-1}$. Since $x^p-1 \equiv x-1 \not \equiv 0 \pmod{p}$, it follows that we only have to show $p^2 \mid x^{3p}-1$. This is again by Exponent Lifting—we have $$\nu_p(x^{3p}-1)=1+\nu_p(x^3-1)\geq 2,$$which yields the desired result. $\blacksquare$

Can someone check this because none of the solutions above are matching this one. Thanks. We will refer to this as the modfrac relation: $$\frac{1}{k} \equiv (-1)^{p-1}\cdot \frac{1}{p}\binom{p}{k} \pmod{p}.$$It suffices to show \begin{align*}\frac{(p-1)!}{p} \left[\binom{p}{1}x + \binom{p}{2}x^2 + \binom{p}{3}x^3 + \cdots + \binom{p}{p-1}x^{p-1}\right] \equiv 0 \pmod{p} \\ x - x^2 + x^3 + \cdots - x^{p-1} \equiv 0 \pmod{p}\end{align*}by the modfrac relation. So we know \begin{align*}x + x^3 + \cdots + x^{p-2} \equiv x^2 + x^4 + \cdots + x^{p-1} \pmod{p} \\ (x-1)(x)(1 + x^2 + \cdots + x^{p-3}) \equiv 0 \pmod{p} \\ (1 + x^2 + \cdots + x^{p-3}) \equiv 0 \pmod{p} \\ \frac{1(1 - (x^2)^{(p-1)/2})}{1-x^2} \equiv 0 \pmod{p} \end{align*}the second to last line coming from the fact that $x^3 - 1 \equiv 0 \pmod{p}$ and $x \not \equiv 0 \pmod{p}$ which is why we can divide by $x$ and $x-1$ and the last line coming from the geometric formula. Hence by Euler totient function we know that \[\frac{1(1 - (x^2)^{(p-1)/2})}{1-x^2} \equiv \frac{1(1 - (x^{p-1}))}{1-x^2} \equiv 0 \pmod{p}\]as desired.

Discard the $(p-1)!$ term, and allow fractions mod $p$. We will use the well known identity that $$\frac{1}{k}\equiv (-1)^{k+1}{p \choose k}\pmod p.$$Then, we have $$\sum_{n=1}^{p-1}\frac{x^n}{n}(-1)^{n-1}\equiv\sum_{n=1}^{p-1}x^n\frac{1}{p}{p \choose k}\pmod p.$$Thus, ti remains to show that $$\sum_{n=1}^{p-1}x^n\frac{1}{p}{p \choose k}\equiv 0\pmod p,$$which is also $$\sum_{n=1}^{p-1}x^n{p \choose k}\equiv 0\pmod {p^2}.$$By binomial theorem, the LHS is $(x+1)^p-1-x^p$, so it suffices to show that $$(x+1)^p\equiv x^p+1\pmod p.$$ This is easily verifiable by testing all possible cases for $p=3$. Assume $p>3$ from now on. Note that since $p\mid x^3-1$ but $p$ does not divide $ x-1$, we have $p\mid x^2+x+1.$ Thus, $x+1\equiv -x^2\pmod p$, so it suffices to show that $$-x^{2p}\equiv x^p+1\pmod {p^2}$$or $$x^{2p}+x^p+1\equiv 0\pmod{p^2}.$$Since the order of $x$ mod $p$ is 3, $x^p-1$ is not divisible by $p$ since we are assuming $p\neq 3$. Thus, it suffices to show that $$x^{3p}-1\equiv 0\pmod {p^2}.$$Rewrite as $$(x^3)^p-1^p\equiv 0\pmod {p^2}.$$This is just LTE: $$v_p((x^3)^p-1^p)=v_p(x^3-1)+v_p(p)\geq1+1=2,$$so we are done.

Lemma: We have that for a prime $p$ and $1 \leq k \leq p$ that $1/k \equiv (-1)^{k - 1}/p \binom{p}{k} \pmod p$. Multiplying the conclusion by a factor of $p$, one concludes that it suffices to show that the sum $\sum_{k = 1}^{p - 1} \binom{p}{k}x^k = (1 + x)^p - (1 + x^p)$ is precisely $0$ modulo $p^2$. Now using the common fact that if for any arbitrary integers $a, b$ we have that $a \equiv b \pmod{p^n}$, then $a^p \equiv b^p \pmod{p^{n + 1}}$, we have that as $x + 1 \equiv -x^2 \pmod p \implies (1 + x)^p \equiv -x^{2p} \pmod{p^2}$, so at present it suffices to show $x^{2p} + x^p + 1 \equiv 0 \pmod{p^2}$. Using now the problem condition, we have that $p \mid x^3 - 1$, and in particular, we obtain $p^2 \mid x^{3p} - 1$, and so it once more and finally suffices to show $p^2 \nmid x^p - 1$. However this is trivial, as $v_p(x^p - 1) = v_p(p) + v_p(x - 1) = 1$, so that $p^2 \nmid x^p - 1$, as desired. $\blacksquare$

Used the 10% hint on ARCH. Use the harmonic modulo $p$ trick, so that now we want to prove that $p^2$ divides \[ \sum_{k=1}^{p-1} \binom pk x^k=(x+1)^p-x^p-1. \]Since $x$ is a primitive cube root of unity, $p$ must be $1$ mod $3$. Now use LTE to see that \[ \nu_p\left((x+1)^p-(x+1-(x^2+x+1))^p\right)=\nu_p(x^2+x+1)+\nu_p(p)\ge 2, \]where $p\nmid x+1$ because $p>3$. Therefore, we want to prove that $p^2$ divides \[ -x^{2p}-x^p-1. \]Noting that $p\nmid x^p-1$, it suffices to prove that $p^2$ divides $x^{3p}-1$. However, this is a direct application of LTE as $x^{3p}-1=(x^3)^p-1$, so we win. $\blacksquare$

By using Wilson's Theorem, note that it suffices to prove that, \[ \sum_{i=1}^{p-1} \dfrac{(-1)^ix^i}{i} \equiv 0 \pmod{p}. \] We know that $\dfrac{(-1)^{k-1}p}{k} \equiv \dbinom{p}{k} \pmod{p^2}$. Now note that proving $\displaystyle\sum_{i=1}^{p-1} \dfrac{(-1)^ix^i}{i} \equiv 0 \pmod{p}$ is equivalent to proving $\displaystyle\sum_{i=1}^{p-1} \dfrac{p(-1)^ix^i}{i} \equiv 0 \pmod{p^2}$. Now, we have, \begin{align*} \sum_{i=1}^{p-1} \dfrac{p(-1)^ix^i}{i} &= - \sum_{i=1}^{p-1} \dfrac{(-1)^{i-1}p}{i}\cdot x^i \\ &\equiv -\sum_{i=1}^{p-1} \binom{p}{i}x^i \\ &= - \left(\sum_{i=0}^{p} \binom{p}{i}x^i - 1 - x^p \right) \\ &= x^p + 1 - \sum_{i=0}^{p} \binom{p}{i}x^i\\ &\equiv x^p + 1 - (1+x)^p\\ &\equiv x^p + 1 +x^{2p}\\ &= \dfrac{x^{3p} - 1}{x^p-1} \pmod{p^2} .\end{align*} Firstly note that $x^p -1 \equiv x - 1 \not\equiv 0 \pmod{p}$. Now note that if we can show that $\nu_{p}\left(\dfrac{x^{3p} - 1}{x^p-1}\right) \ge 2$, then we are done. By LTE, $\nu_{p}\left(\dfrac{x^{3p} - 1}{x^p-1}\right) = \nu_{p}(x^{3p} -1) - \nu_{p}(x^p - 1) = (\nu_{p}(x^3 - 1) + \nu_{p}(p)) - 0 \ge 2$ and we are done.

$\textcolor{blue}{Lemma.}$ $$\frac{1}{k}\equiv \frac{(-1)^{k-1}}{p}\binom{p}{k}\pmod {p^2}$$ By the lemma and Wilson´s Theorem it suffices to prove $$\binom{p}{1}x^1+\binom{p}{2}x^2+\cdots \binom{p}{p-1}x^{p-1}\equiv 0\pmod {p^2}$$Which is equivalent to $$(x+1)^p\equiv x^p+1\pmod{p^2}$$ then it suffices to show $p^2\mid f(x)=(x+1)^p-x^p-1$, we are going to prove that $f(x)$ has $\zeta_3$ as a double root, with this we finish as any polynomial that has $\zeta_p$ as a root is divisible by $x^{p-1}+\cdots+x+1$; as $\operatorname{Ord}_p(x)=3$ then $6\mid p-1$; note that $$f(\zeta_3)=(1+\zeta_3)^p-\zeta_3^p-1=1+\zeta_3-\zeta_3-1=0 \hspace{0.5cm}\text{and}\hspace{0.5cm} f^{\prime}(\zeta_3)=p(1+\zeta_3)^{p-1}-p\zeta_3^{p-1}=p\cdot 1-p\cdot 1=0$$(we used the fact that $1+\zeta_3$ is a 6-th rooth of the unity) And we are done.

can't we use p adic analysis here/?/???

$ $

Since $x^3 \equiv 1$ (mod $p$) and $x \not\equiv 1$ (mod $p$), we get that $ord_p(x)=3$, so $3 \mid p-1$. Since $p$ is odd, this implies that $6 \mid p-1$. Now, $x - \frac {x^2}{2} + \frac {x^3}{3} - \cdots - \frac {x^{p - 1}}{p - 1} $ $= \sum_{i=0}^{\frac{p-1}{6}-1} \left( \frac{x^{6i+1}}{6i+1}-\frac{x^{6i+2}}{6i+2}+\frac{x^{6i+3}}{6i+3}-\frac{x^{6i+4}}{6i+4}+\frac{x^{6i+5}}{6i+5}-\frac{x^{6i+6}}{6i+6} \right) \equiv \sum_{i=0}^{\frac{p-1}{6}-1} \left( \frac{x}{6i+1}+\frac{x+1}{6i+2}+\frac{1}{6i+3}-\frac{x}{6i+4}-\frac{x+1}{6i+5}-\frac{1}{6i+6} \right) $ $= x\sum_{i=0}^{\frac{p-1}{6}-1} \left( \frac{1}{6i+1}+\frac{1}{6i+2}-\frac{1}{6i+4}-\frac{1}{6i+5} \right) + \sum_{i=0}^{\frac{p-1}{6}-1} \left( \frac{1}{6i+2}+\frac{1}{6i+3}-\frac{1}{6i+5}-\frac{1}{6i+6} \right) $ $\equiv \frac{x}{p}\sum_{i=0}^{\frac{p-1}{6}-1} \left( \binom{p}{6i+1}-\binom{p}{6i+2}+\binom{p}{6i+4}-\binom{p}{6i+5} \right) + \frac{1}{p} \sum_{i=0}^{\frac{p-1}{6}-1} \left( -\binom{p}{6i+2}+\binom{p}{6i+3}-\binom{p}{6i+5}+\binom{p}{6i+6} \right) $ $= \frac{x}{p} \left( \sum_{i=0}^{\frac{p-1}{3}-1} \binom{p}{3i+1} - \sum_{i=0}^{\frac{p-1}{3}-1} \binom{p}{3i+2} \right) + \frac{1}{p} \left( \sum_{i=1}^{\frac{p-1}{3}} \binom{p}{3i} - \sum_{i=0}^{\frac{p-1}{3}-1} \binom{p}{3i+2} \right) = \frac{x}{p} \left( \frac{2^p-2}{3} - \frac{2^p-2}{3} \right) + \frac{1}{p} \left( \frac{2^p-2}{3} - \frac{2^p-2}{3} \right) =0 $ (mod $p$).

Recall the well-known identity $$\frac{1}{k} \equiv (-1)^{k-1} \cdot \frac{1}{p} \binom{p}{k} \pmod p.$$It thus suffices to show that $$\sum_{k=1}^{p-1} \binom{p}{k} x^k \equiv 0 \pmod {p^2}.$$By the Binomial Theorem, this reduces to $$(x+1)^p \equiv x^p+1 \pmod{p^2}.$$Now, notice that by LTE $v_p((x+1)^p+(x^2)^p) \geq 2,$ so $(x+1)^p \equiv -(x^2)^p \pmod{p^2}.$ Hence, it suffices to show that $$x^{2p} + x^p+1 \equiv 0 \pmod{p^2}.$$Clearly, $x^p-1 \not \equiv 0 \pmod p.$ Meanwhile, by LTE, $$v_p(x^{3p}-1) = v_p(x^3-1)+v_p(p) \geq 2,$$and thus the desired result follows. QED