Solution:

There is another solution that avoids getting a closed form. I used exactly the notation of the previous post in defining $T_a$, so I won't redefine it here. We will solve the problem by proving the following two identities: \[ T_a \equiv T_{1-a} \pmod{p}, \] \[ T_{a^2} \equiv (T_a + T_{-a}) + a(T_a - T_{-a}) \pmod{p}. \] For the first identity, the coefficient of $a^k$ in $T_{1-a}$ can be computed with the binomial theorem. For $1 \leq k \leq p-1$, we get \begin{align*} [a^k] T_{1-a} &\equiv (-1)^k \sum_{i=1}^{p-1} \binom{i}{k} i^{-1} \pmod{p}, \\ &\equiv (-1)^k \sum_{i=1}^{p-1} \binom{i-1}{k-1} k^{-1} \pmod{p}, \\ &\equiv (-1)^k k^{-1} \binom{p-1}{k} \pmod{p}. \end{align*} Since we can easily show that $\binom{p-1}{k} \equiv (-1)^k \pmod{p}$, we get the coefficient of $a^k$ is equivalent to $k^{-1}$ mod $p$. For $k = 0$, we quickly see it is the sum of the inverses of all nonzero residues mod $p$, which is zero. The claim that $T_{1-a}$ is equivalent to $T_a$ mod $p$ is now evident. For the second identity, notice that \[ T_{a^2} 2^{-1} \equiv a^2 \cdot 2^{-1} + a^4 \cdot 4^{-1} + \cdots + a^{2(p-1)} \cdot [2(p-1)]^{-1} \pmod{p}, \] Let $E$ be the first half of the terms on the right side, up to $a^{p-1} (p-1)^{-1}$, and let $O$ be the second half of the terms. Notice that $T_a + T_{-a} \equiv 2E \pmod{p}$, since $E$ includes only the terms of even exponent in $T_a$. Similarly, since $a^{p+k} \equiv a^{k+1} \pmod{p}$, $O$ includes only the terms of odd exponent in $T_a$ except all multiplied by $a$. Therefore, $a(T_a - T_{-a}) \equiv 2O \pmod{p}$. Our claim for the formula of $T_{a^2}$ follows immediately. Now apply the first identity for $a = 3$ and the second identity for $a = 2$. We get that \[ T_3 + T_4 - 3T_2 \equiv T_{-2} + (T_2 + T_{-2}) + 2(T_2 - T_{-2}) - 3T_2 \equiv 0 \pmod{p}, \] as desired.

I would just like to point out that there have been more difficult problems using this trick (of moving between $\pmod{p}$ and $\pmod{p^2}$) before: for instance, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=289232]2009 ELMO[/url] wrote: 6. (John Berman) Let $ p$ be an odd prime and $ x$ be an integer such that $ p \mid x^3 - 1$ but $ p \nmid x - 1$. Prove that \[ p \mid (p - 1)!\left(x - \frac {x^2}{2} + \frac {x^3}{3} - \cdots - \frac {x^{p - 1}}{p - 1}\right).\] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=487132&]2011 ELMO Shortlist[/url] wrote: Let $p\ge5$ be a prime. Show that \[\sum_{k=0}^{(p-1)/2}\binom{p}{k}3^k\equiv 2^p - 1\pmod{p^2}.\]

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3557450&sid=1d0ef5e8daa02762cfdd2f824a8001f3#p3557450 is a much harder problem that essentially uses this problem to complete the solution after the initial simplification of the expression.

Hi guys! The 2015WOOTer has just learned p-adic valuations, and would like to test his knowledge. However, he isn't sure if his solution works, so can someone critique it?

Solution : First we need to prove that $S_4 - 3S_2 +S_{-2}=0 mod p$. We get $S_4 = \sum_{i=1}^{(p-1)/2}4^i(\frac{1}{i} + \frac{1}{(p-1)/2+i})$, $2S_2 = \sum_{i=1}^{(p-1)/2}\frac{4^i}{i} + \sum_{i=1}^{(p-1)/2}\frac{4^i}{2i-1}$ and so $S_4-2S_2 = \sum_{i=1}^{(p-1)/2}4^i(\frac{1}{(p-1)/2+i}- \frac{1}{2i-1}) = \sum_{i=1}^{(p-1)/2}4^i\frac{1}{2i-1}=2\sum_{i=1}^{(p-1)/2}\frac{2^{2i-1}}{2i-1}$ and so $S_4-3S_2 = S_{-2} mod p$. Next we need to prove that $S_k = S_{1-k} mod p$. We get $S_{1-k} = \sum_{i=1}^{p-1}\frac{\sum_{j=0}^ik^jC_i^j}{i} = \sum_{i=1}^{p-1}(1/i+\frac{\sum_{j=1}^ik^jC_i^j}{i}) = \sum_{i=1}^{p-1}\frac{\sum_{j=1}^ik^jC_i^j}{i} = \sum_{j=1}^{p-1}k^j\sum_{i=j}^{p-1}\frac{(-1)^jC_i^j}{i}$, so we only need to prove that $\sum_{i=j}^{p-1}\frac{(-1)^jC_i^j}{i} = 1/j mod p$. This is equivalent to $\sum_{i=j}^{p-2}C_i^j = (-1)^j mod p$ which is well known fact. So we get that $S_3+S_4-3S_2 = S_3 - S_{-2} = 0 mod p$. $\Box$

How this is N7 ? I solved this is 7 minutes lololol (this is clearly the easiest problem in the entire shortlist)

Note that $S_3=\sum_{j=1}^{p-1}\frac{3^j}{j}=\sum_{j=1}^{p-1}\frac{1}{j}\sum_{k=0}^j\binom{j}{k}2^k=\sum_{j=1}^{p-1}\frac{1}{j}+\sum_{k=1}^{p-1}\sum_{j=k}^{p-1}\frac{1}{k}\binom{j-1}{k-1}2^k\equiv \sum_{k=1}^{p-1}\frac{2^k}{k}\sum_{j=k}^{p-1}\binom{j-1}{k-1}=\sum_{k=1}^{p-1}\frac{2^k}{k}\binom{p-1}{k}\equiv \sum_{k=1}^{p-1}\frac{(-2)^k}{k}$. So, $S_3-S_2=\sum_{k=1}^{p-1}\frac{(-2)^k-2^k}{k}=\sum_{k=1}^{\frac{p-1}{2}}\frac{-2^{2k}}{2k-1}$. Also, $S_4=\sum_{k=1}^{\frac{p-1}{2}}2^{2k}\left(\frac{1}{k}+\frac{1}{\frac{p-1}{2}+k}\right)\equiv\sum_{k=1}^{\frac{p-1}{2}}\frac{2^{2k+1}}{2k}+\sum_{k=1}^{\frac{p-1}{2}}\frac{2*2^{2k}}{2k-1}$. This means $S_3-S_2+S_4=\sum_{k=1}^{\frac{p-1}{2}}\frac{2^{2k+1}}{2k}+\sum_{k=1}^{\frac{p-1}{2}}\frac{2^{2k}}{2k-1}=\sum_{k=1}^{p-1}\frac{2*2^k}{k}=2S_2$, and $S_3+S_4-3S_2\equiv 0$ as desired.

Easiest N7 ever (especially if you have seen ELMO 2009 P6 before). Here's my solution: Note that, for any $1 \leq i \leq p-1$, we have $$\frac{1}{i}=\frac{(p-i)!(i-1)!}{i!(p-i)!}=\frac{(i-1)!(p-(p-1))(p-(p-2)) \dots (p-i)}{i!(p-i)!} \equiv \frac{(-1)^{p-i} (p-1)!}{i!(p-i)!}=\frac{(-1)^{i-1}}{p} \binom{p}{i} \pmod{p}$$So, using Binomial Theorem, we get $$S_a \equiv \sum^{p-1}_{j=1} \left(\frac{(-1)^{j-1}}{p} \binom{p}{j} \cdot a^j \right) \equiv \frac{1}{p} ((a-1)^p-(a^p-1)) \pmod{p}$$On expanding, this translates to $$S_3+S_4-3S_2 \equiv \frac{4}{p} ((2^p-1)-4^{p-1}) \pmod{p}$$So we just need to show that $$4^{p-1}+1 \equiv 2^p \pmod{p^2} \Longleftrightarrow (2^{p-1}-1)^2 \equiv 0 \pmod{p^2}$$which is true by FLT. $\blacksquare$

Not so hard after ELMO 2009 P6, as math_pi_rate said. Anyway, just for storage. Solution. We will use $a\equiv b$ to note $a\equiv b \pmod{p}$ wherever required in the rest of the proof. We present the following lemma which basically kills the problem. Lemma. We have $$\frac{1}{\ell}\equiv \frac{(-1)^{\ell-1}}{p}\binom{p}{\ell} \pmod{p}$$Proof. Notice that \begin{align*} \binom{p}{\ell} \frac{(-1)^{\ell-1}}{p} &= \left(\frac{p(p-1)\ldots (p-\ell+1)}{\ell(\ell-1)\ldots 1}\right) \cdot \frac{(-1)^{\ell-1}}{p} \\ &\equiv \frac{(-1)(-2)\ldots (-(\ell-1))}{\ell (\ell-1)!} \cdot (-1)^{\ell-1} \\ &= \frac{1}{\ell}\cdot\frac{(-1)^{\ell-1}(\ell-1)!}{(\ell-1)!}\cdot (-1)^{\ell-1} \\ &= \frac{1}{\ell} \\ \end{align*}as desired. $\square$ It suffices to show $$(S_3+S_4-3S_2) \equiv 0 \pmod{p} \iff p\left(\frac{S_3}{p}+\frac{S_4}{p}-\frac{3S_2}{p}\right)\equiv 0 \pmod{p}$$While using above Lemma and Binomial Theorem, note the following, : \begin{align*} p\left(\frac{S_3}{p}+\frac{S_4}{p}-\frac{3S_2}{p}\right) &= \frac{1}{p}\left\{p\left[\frac{3}{1}+\frac{3^2}{2}+\ldots+\frac{3^{p-1}}{p-1}\right]+p\left[\frac{4}{1}+\frac{4^2}{2}+\ldots+\frac{4^{p-1}}{p-1}\right]-3p\left[\frac{2}{1}+\frac{2^2}{2}+\ldots+\frac{2^{p-1}}{p-1}\right]\right\} \\ &\equiv \frac{1}{p}\left\{\cancel{p}\left[\frac{3}{\cancel{p}}\binom{p}{1}-\frac{3^2}{\cancel{p}}\binom{p}{2}+\ldots-\frac{3^{p-1}}{\cancel{p}}\binom{p}{p-1}\right]+\cancel{p}\left[\frac{4}{\cancel{p}}\binom{p}{1}-\frac{4^2}{\cancel{p}}\binom{p}{2}+\ldots-\frac{4^{p-1}}{\cancel{p}}\binom{p}{p-1}\right]-3\cancel{p}\left[\frac{2}{\cancel{p}}\binom{p}{1}-\frac{2^2}{\cancel{p}}\binom{p}{2}+\ldots-\frac{2^{p-1}}{\cancel{p}}\binom{p}{p-1}\right]\right\}\\ &=\frac{1}{p}\{2^p+1-3^p+3^p+1-4^p-3(1+1-2^p)\} \\ &=\frac{1}{p}\{4\cdot(2^{p})-(2^p)^2-4\}\\ &=\frac{1}{p}[-(2^p-2)^2] \equiv \frac{1}{p}\cdot -(pk)^2=-pk^2\equiv 0\\ \end{align*}where $-(2^p-2)^2\equiv -(pk)^2$ follows from Fermat Little Theorem, finishing the problem. $\blacksquare$

By the harmonic modulo $p$ trick, \[S_a\equiv \tfrac 1p\sum_{j=1}^{p-1}a^j(-1)^{j-1}\binom{p}{j}\equiv -\tfrac 1p((1-a)^p-1+a^p)\pmod{p}.\]Thus it is sufficient to show \[p^2\mid (1-3)^p-1+3^p+(1-4)^p-1+4^p-3\cdot (1-2)^p+3\cdot1-3\cdot2^p=\]\[-2^p-1+3^p-3^p-1+4^p+3+3-3\cdot2^p=\]\[4^p-4\cdot2^p+4=(2^p-2)^2.\]Since $p\mid 2^p-2$, we are done.

This dies to harmonic modulo $p$ lemma as many above as mentioned but posting for storage : We see that $\dfrac{1}{k} \equiv (-1)^{k+1} \cdot \dfrac{1}{p}\dbinom{p}{k} \pmod{p}$ is basically harmonic modulo $p$ lemma, so $S_x \equiv \dfrac{1}{p} \sum\limits_{k=1}^{p-1} x^k (-1)^{k-1} \dbinom{p}{k} \equiv \dfrac{-1}{p}((1-x)^p + x^p-1)$, and $S_3 + S_4 - 3S_2$ simply boils down to $(2^p-2)^2$, and we see that $p^2 \lvert (2^p-2)^2$ using Fermat's Theorem and hence $p \lvert S_3 + s_4 - 3S_2$ as desired.

The following lemma basically kills the problem. Lemma. Let $p$ be an odd prime, suppose $1\leq k\leq p-1$, then $$k^{-1}\equiv (-1)^{k+1}\frac{1}{p}\binom{p}{k}\pmod p$$Proof. This follows from $$\frac{k}{p}\binom{p}{k}=\binom{p-1}{k-1}\equiv (-1)^{k-1}\pmod p$$$\blacksquare$ Now if $p=3$ it is obvious, otherwise assume $p\geq 5$, we have $$S_3+S_4-3S_2=\sum_{j=1}^{p-1}\frac{3^j+4^j-3\cdot 2^j}{j}=\frac{\displaystyle\sum_{j=1}^{p-1}\frac{(p-1)!}{j}(3^j+4^j-3\cdot 2^j)}{(p-1)!}$$Notice that $((p-1)!,p)=1$ and $(p-1)!\equiv -1\pmod p$ by Wilson's theorem. Therefore, it suffices to show $$\sum_{j=1}^{p-1}(3^j+4^j-3\cdot 2^j)j^{-1}\equiv 0\pmod p\hspace{20pt}(1)$$From the lemma we have $$\sum_{j=1}^{p-1}a^jj^{-1}\equiv -\sum_{j=1}^{p-1}\binom{p}{k}(-a)^k=\frac{a^p-(a-1)^p-1}{p}\pmod p$$Therefore, left hand side of $(1)$ is congruent to $$\frac{3^p-2^p-1+4^p-3^p-1-3(2^p-1^p-1)}{p}=\frac{4^p-4\cdot 2^p+4}{p}=\frac{4(2^{p-1}-1)^2}{p}\equiv 0\pmod p$$as desired.

Easy for N7(posting for storage): We'll show that $p$ divides the numerator of $S_n$ By some manipulation we get :- $(-1)^{2 (p-1)} \cdot \sum_{i=1}^{p-1} \frac {x^i}{i} \equiv - $$ \sum_{1\le i\le p-1}\frac{(-1)^{i-1}}{i}x^i$ Now,we will show that $p$ divides the numerator of $x - \frac{x^2}{2} + \frac{x^3}{3} ... -\frac{x^{p-1}}{p-1}$. Now observe that $\textstyle \frac1p \binom{p}{i}\equiv (-1)^{i-1}/i\pmod{p}$. Using this, the given sum computes, in modulo $p$, to $ \sum_{1\le i\le p-1}\frac{(-1)^{i-1}}{i}x^i = \frac1p\sum_{1\le i\le p-1}\binom{p}{i}x^i = \frac{(x+1)^p-x^p-1}{p}. $So, it suffices to show that if $p\mid x^2+x+1$ then $p^2\mid (x+1)^p-x^p-1$. By Fermat's theorem $(x+1)^p-x^p-1\equiv 0\pmod{p}$, so set $(x+1)^p-x^p-1=pQ(x)$ for some $Q(x)\in\mathbb{Z}[x]$. Furthermore, $x+1\equiv -x^2\pmod{x^2+x+1}$, forcing $ (x+1)^p-x^p-1\equiv -x^{2p}-x^p-1\equiv 0\pmod{x^2+x+1} $ as required. Hence $\sum_{i=1}^{p-1} \frac {x^i}{i} \equiv $$- \sum_{1\le i\le p-1}\frac{(-1)^{i-1}}{i}x^i \equiv 0 \pmod p $

Lemma:$\frac{1}{k} \equiv \frac {(-1)^{k-1}}{p} \binom pk (mod p)$ $$\frac{m}{n}=S_3 + S_4 - 3S_2 = \sum^{p-1}_{j=1} \frac{(3^j+4^j-3 \cdot 2^j)}{j} \equiv \sum^{p-1}_{j=1} \frac{(3^j+4^j-3 \cdot 2^j) \cdot (-1)^{j-1} \cdot \binom pj}{p} \equiv \frac{4 \cdot 2^p -4^p -4}{p} \equiv - \frac{(2^p-2)^2}{p} \equiv 0 $$mod($p$) $\implies$ $p|m$

We use the well-known identity $$\frac{1}{k} \equiv (-1)^{k-1}\cdot\frac{1}{p}\binom{p}{k} \pmod{p},$$which actually just kills the problem lol. Using the identity, we have \begin{align*} S_3+S_4-3S_2&=\sum_{j=1}^{p-1} \frac{3^j+4^j-3\cdot 2^j}{j}\\ &\equiv\frac{1}{p}\left(3\sum_{j=1}^{p-1} (-2)^j\binom{p}{k}-\sum_{j=1}^{p-1} (-3)^j\binom{p}{k}-\sum_{j=1}^{p-1} (-4)^j \binom{p}{k}\right):=\frac{N}{p} \pmod{p}, \end{align*}so we want to show that $p^2 \mid N$. By the binomial theorem, \begin{align*} N&=3(1-2)^p-3-3(-2)^p-(1-3)^p+1+(-3)^p-(1-4)^p+1+(-4)^p\\ &=-3-3+3\cdot2^p+2^p+1-3^p+3^p+1-4^p\\ &=4^p-4\cdot 2^p+4=(2^p-2)^2. \end{align*}Since $p \mid 2^p-2$, $p^2 \mid N$, so we're done. $\blacksquare$

The central lemma we will use here is the fact that \[ \frac{1}{k} \equiv (-1)^{k-1}\cdot\frac{1}{p}\binom{p}{k} \pmod{p}. \]We apply this to obtain \[ S_3 + S_4 - 3S_2 = \sum^{p-1}_{j=1} \frac{(3^j+4^j-3 \cdot 2^j)}{j} \equiv \frac{1}{p} \sum^{p-1}_{j=1} (3^j+4^j-3 \cdot 2^j)(-1)^{j-1}\binom{p}{j} \pmod{p}. \] Using the Binomial Theorem, the sum $\sum^{p-1}_{j=1} (3^j+4^j-3 \cdot 2^j)(-1)^{j-1}\binom{p}{j}$ can be evaluated as \[ \sum^{p-1}_{j=1} (3^j+4^j-3 \cdot 2^j)(-1)^{j-1}\binom{p}{j} = 4^p-4 \cdot 2^p+4=(2^p-2)^2. \]Hence, \[ S_3 + S_4 - 3S_2 \equiv \frac{(2^p-2)^2}{p} \equiv 0 \pmod{p}, \]and we are done.

Denote $f(x) = \sum_{i=1}^{p-1} \frac {x^i} {i} = S_x$ Lemma: $f(x + 1) \equiv f(-x) \mod{p} $ Proof: \begin{align*} f(x + 1) = \sum_{i=1}^{p-1} \frac {(x + 1)^i} {i} &= \sum_{i=1}^{p-1} \frac {\sum_{j=0}^i x^j \binom{i}{j}} {i} = \sum_{i=1}^{p-1} \left ( x^i \left ( \sum_{j=1}^{p-1} \frac {\binom{j}{i}} {j} \right) \right ) \\ &= \sum_{i=1}^{p-1} \left ( x^i \left ( \sum_{j=1}^{p-1} \frac { (j-1) \cdot \dots \cdot (j - (i-1))} {i!} \right) \right ) \\ &= \sum_{i=1}^{p-1} \left ( \frac {x^i} {i!} \left ( \sum_{j=1}^{p-1} j^{i-1} - (1 + \dots + (i-1))\sum_{j=1}^{p-1} j^{i-2} + \dots + \sum_{j=1}^{p-1} (-1)^{i-1} (i-1)! \right) \right)\\ &\equiv \sum_{i=1}^{p-1} \left ( \frac {x^i} {i!} \left ( 0 - (1 + \dots + (i-1)) \cdot 0 + \dots + (p - 1) \cdot (-1)^{i-1} (i-1)! \right) \right) \\ &\equiv \sum_{i=1}^{p-1} \left ( \frac {x^i} {i!} \left ((-1)^{i} (i-1)! \right) \right) = \sum_{i=1}^{p-1} \frac { (-x)^i } {i} = f (-x) \\ \end{align*} So: \begin{align*} S_3 + S_4 - 3 S_2 &= f(3) + f(4) - 3 f(2) \equiv f(-2) + f(4) - 3 f(2) \mod{p} \\ f(4) + ( f(-2) - 3 f(2) ) &= \sum_{k = 1}^{p - 1} \frac { 2^{2k} } {k} + \sum_{k = 1}^{p - 1} \frac { (-2)^{2k} } {k} - 3 \sum_{k = 1}^{p - 1} \frac { 2^k } {k}\\ &= \sum_{k = 1}^{\frac{p - 1} {2}} \left ( \frac {2^{2k}} {k} + \frac {2^{ 2(k + \frac {p-1} {2} ) } } {k + \frac {p-1} {2} } \right ) + \sum_{k = 1}^{\frac{p - 1} {2}} \frac { (-2) ^ {2k} } {2k} + \sum_{k = 1}^{\frac{p - 1} {2}} \frac { (-2)^{ 2k - 1} } { 2k - 1} - 3 \sum_{k = 1}^{\frac{p - 1} {2}} \frac {2^{2k}} {2k} - 3 \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2 ^ {2k - 1}} {2k - 1} \\ &\equiv \sum_{k = 1}^{\frac{p - 1} {2}} \left ( \frac {2^{2k}} {k} + \frac {2^{2k} } {k + \frac {-1} {2} } \right ) + \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k} } {2k} - 3 \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k} } {2k} - \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k - 1} } {2k - 1} - 3 \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k - 1} } {2k - 1}\\ &\equiv \sum_{k = 1}^{\frac{p - 1} {2}} \left ( \frac {2^{2k}} {k} + \frac {2^{2k} } {\frac { 2k - 1} {2} } \right ) - 2 \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k} } {2k} - 4 \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k - 1} } {2k - 1}\\ &\equiv \sum_{k = 1}^{\frac{p - 1} {2}} \left ( \frac {2^{2k}} {k} + \frac {2^{2k + 1} } {2k -1} \right ) - \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k} } {k} - \sum_{k = 1}^{\frac{p - 1} {2}} \frac { 2^ {2k + 1} } {2k - 1}\\ &\equiv 0 \mod{p} \end{align*}So $p$ devides $m$!

Same as the others, but I don't think they've provided a proof to the lemma. https://paste.pics/c60a9e029d9d776702f120efd4c82671

Use the harmonic modulo $p$ trick to have the sum instead look like \[-\frac1p\sum_{j=1}^{p-1}\binom pj (-a)^j=(1-a)^p+a^p-1.\]Then \[p(S_3+S_4-3S_2)\equiv 4(4^{p-1}-2^p+1)\pmod{p^2}.\]But the second factor is just $(2^{p-1}-1)^2$, which is clearly divisible by $p^2$, so we are done. $\blacksquare$

Recall the harmonic modulo $p$ trick, stating that for any integer $k=1, 2, \cdots, p-1,$ we have $$\frac{1}{k} \equiv (-1)^{k-1} \cdot \frac{1}{p} \binom{p}{k} \pmod p.$$Therefore, $$S_a \equiv \sum_{j=1}^{p-1} \frac{a^j (-1)^{j-1}}{p} \binom{p}{j} \equiv -\frac{1}{p} \sum_{j=1}^{p-1} (-a)^j \binom{p}{j} \equiv \frac{1}{p} \left( (a-1)^p-a^p+1 \right) \pmod p.$$Therefore, $S_3+S_4-3S_2 \equiv \frac{1}{p} \left( 2^p-3^p+1+3^p-4^p+1-3 \cdot 1 + 3\cdot 2^p-3 \right) \equiv -\frac{1}{p} (2^p-2)^2 \pmod p,$ so it suffices to show that $2^p-2 \equiv 0 \pmod p,$ which is true by FLT. QED Remark: This was made trivial by ELMO 2009 P6 LOLL