Let $p$ be an odd prime. For each integer $a$, set
$$S_a=\sum_{k=1}^{p-1}\frac{a^k}{k} .$$
Write $S_4+S_3-3S_2$ in the form $$S_4+S_3-3S_2=\frac{m}{n}$$ where $n,m\in\mathbb Z$ satisfy $\gcd(n,m)=1$.
How to show that $p $ divides $ m$ ?
Tried looking at first and last $k$th elements:
$$\sum_{k=1}^{p-1}\frac{a^k}{k}=\sum_{k=1}^{\frac{p-1}{2}}\frac{a^k(p-k)+a^{p-k}k}{k(p-k)}=\sum_{k=1}^{\frac{p-1}{2}}\frac{a^kp-a^kk+a^{p-k}k}{k(p-k)}=\frac m n$$
If it were $a^k=1$, we'd be done (We'd have harmonic sum). But the problem is $a^k=4^k+3^k-3\cdot2^k$ and the above observation can't help with that.
Grouping terms of sum seems a dead end. For example,
$$\sum_{k=1}^{p-1}\frac{a^k}{k}=\frac{\sum_{k=1}^{p-1}a^k\frac{(p-1)!}{k}}{(p-1)!}$$
And looking modulo $p$ terms in the numerator, doesn't yield anything nice.
Even though $p=5$ yields false hope $4+1+1+4=0+0=0$ of grouping $1.$,$2.$ and $3.$,$4.$,
already, for example, $p=11$ gives $10+ 10+ 7+ 2+ 10+ 7+ 1+ 9+ 0+ 10=0$, a sequence without a pattern, so I'm not sure how to go on proving they will all sum to $0$ modulo $p$.
First and last term (Euler's theorem) are evidently $=1=p-1$ modulo $p$, but I'm not sure how the terms in between can be handled.
Nice question! As with Pascal's letter, this could be much shorter if I had more time, but it is what it is. Chances are you can skip a lot of it.
1. Introduction to $p$-integers
Let me first build up a proper language for "congruences between rational numbers modulo powers of $p$". This is folklore on olympiads and in AoPS, and follows easily from some abstract algebra, but since it is much more often used than explained, I will have to include some proofs.
Fix a prime $p$. We do not require $p$ to be odd, unless we say so.
I will use the concept of a $p$-adic valuation of a rational number; this is defined in https://mathoverflow.net/questions/331091/reciprocal-sum-of-binomials-and-divisibility-by-3/331850#331850 . Let me quote Proposition 1 from loc. cit.:
Now, let $\mathbb{Z}_{\left( p\right) }$ be the set of all rational numbers $r\in\mathbb{Q}$ satisfying $v_{p}\left( r\right) \geq0$. The elements of $\mathbb{Z}_{\left( p\right) }$ are called the $p$-integers (not to be mistaken for the $p$-adic integers, whose set is denoted by $\mathbb{Z}_{p}$ without the parentheses in the subscript). The following is well-known:
Proof of Proposition 2. Each integer $r$ satisfies $v_{p}\left( r\right) \geq0$ and thus $r\in\mathbb{Z}_{\left( p\right) }$ (by the definition of $\mathbb{Z}_{\left( p\right) }$). In other words, the set $\mathbb{Z} _{\left( p\right) }$ contains all integers. Thus, this set $\mathbb{Z} _{\left( p\right) }$ contains $0$ and $1$ and $-1$. In other words, $0\in\mathbb{Z}_{\left( p\right) }$ and $1\in\mathbb{Z}_{\left( p\right) }$ and $-1\in\mathbb{Z}_{\left( p\right) }$.
Let $a,b\in\mathbb{Z}_{\left( p\right) }$. From $a\in\mathbb{Z}_{\left( p\right) }$, we obtain $a\in\mathbb{Q}$ and $v_{p}\left( a\right) \geq0$ (by the definition of $\mathbb{Z}_{\left( p\right) }$). Likewise, $b\in\mathbb{Q}$ and $v_{p}\left( b\right) \geq0$. Now, Proposition 1 (a) yields $v_{p}\left( ab\right) =\underbrace{v_{p}\left( a\right) }_{\geq0}+\underbrace{v_{p}\left( b\right) }_{\geq0}\geq0+0=0$. Hence, $ab\in\mathbb{Z}_{\left( p\right) }$ (by the definition of $\mathbb{Z} _{\left( p\right) }$). Moreover, $\min\left\{ v_{p}\left( a\right) ,v_{p}\left( b\right) \right\} $ must be one of the two numbers $v_{p}\left( a\right) $ and $v_{p}\left( b\right) $, and thus is $\geq0$ (since both numbers $v_{p}\left( a\right) $ and $v_{p}\left( b\right) $ are $\geq0$ (because $v_{p}\left( a\right) \geq0$ and $v_{p}\left( b\right) \geq0$)). In other words, $\min\left\{ v_{p}\left( a\right) ,v_{p}\left( b\right) \right\} \geq0$; thus, Proposition 1 (b) yields $v_{p}\left( a+b\right) \geq\min\left\{ v_{p}\left( a\right) ,v_{p}\left( b\right) \right\} \geq0$. Hence, $a+b\in\mathbb{Z}_{\left( p\right) }$ (by the definition of $\mathbb{Z}_{\left( p\right) }$).
Now, forget that we fixed $a,b$. We thus have proven that $a+b\in \mathbb{Z}_{\left( p\right) }$ and $ab\in\mathbb{Z}_{\left( p\right) }$ for all $a,b\in\mathbb{Z}_{\left( p\right) }$. In other words, the subset $\mathbb{Z}_{\left( p\right) }$ of $\mathbb{Q}$ is closed under addition and closed under multiplication.
Moreover, if $a\in\mathbb{Z}_{\left( p\right) }$, then $\left( -1\right) a\in\mathbb{Z}_{\left( p\right) }$ (since $-1\in\mathbb{Z}_{\left( p\right) }$ and $a\in\mathbb{Z}_{\left( p\right) }$, and since $\mathbb{Z}_{\left( p\right) }$ is closed under multiplication) and thus $-a=\left( -1\right) a\in\mathbb{Z}_{\left( p\right) }$. Thus, the set $\mathbb{Z}_{\left( p\right) }$ is closed under additive inverses.
Now, we have shown that the subset $\mathbb{Z}_{\left( p\right) }$ of $\mathbb{Q}$ is closed under addition, closed under multiplication, and closed under additive inverses, and contains $0$ and $1$. Thus, $\mathbb{Z}_{\left( p\right) }$ is a subring of $\mathbb{Q}$. Of course, $\mathbb{Z} \subseteq\mathbb{Z}_{\left( p\right) }$, since $\mathbb{Z}_{\left( p\right) }$ contains all integers. This proves Proposition 2. $\blacksquare$
Thus, in particular, $\mathbb{Z}_{\left( p\right) }$ is a ring. Hence, it makes sense to speak of divisibility and congruence in $\mathbb{Z}_{\left( p\right) }$; these concepts are defined as you would (hopefully) expect:
Divisibility is defined as follows: If $a,b\in\mathbb{Z}_{\left( p\right) }$, then we say that "$a\mid b$ in $\mathbb{Z}_{\left( p\right) }$" if and only if there exists a $c\in\mathbb{Z}_{\left( p\right) }$ such that $b=ac$. When $a$ is nonzero, this is equivalent to saying that $b/a\in\mathbb{Z} _{\left( p\right) }$ (because obviously, if there exists a $c\in \mathbb{Z}_{\left( p\right) }$ such that $b=ac$, then this $c$ must be $b/a$).
Congruence is defined as follows: If $a,b,c\in\mathbb{Z}_{\left( p\right) }$, then we say that "$a\equiv b\operatorname{mod}c$ in $\mathbb{Z}_{\left( p\right) }$" if and only if $c\mid a-b$ in $\mathbb{Z}_{\left( p\right) }$.
For comparison: We shall write "$a\mid b$ in $\mathbb{Z}$" for the usual relation of divisibility of integers, and we shall write "$a\equiv b\operatorname{mod}c$ in $\mathbb{Z}$" for the usual relation of congruence of integers.
In general, if $a$ and $b$ are two integers, then "$a\mid b$ in $\mathbb{Z} _{\left( p\right) }$" is a weaker statement than "$a\mid b$ in $\mathbb{Z}$" (that is, usual divisibility of integers). For example, we have $2\mid3$ in $\mathbb{Z}_{\left( 5\right) }$ (since $3=2\cdot\dfrac{3}{2}$ and $\dfrac {3}{2}\in\mathbb{Z}_{\left( 5\right) }$), but we don't have $2\mid3$ in $\mathbb{Z}$. It turns out that divisibility in $\mathbb{Z}_{\left( p\right) }$ is a fairly rough relation that boils down to an inequality between $p$-adic valuations (the abstract way to say this is "$\mathbb{Z}_{\left( p\right) }$ is a valuation ring"); here is the precise statement:
Proof of Proposition 3. $\Longrightarrow:$ Assume that $a\mid b$ in $\mathbb{Z}_{\left( p\right) }$. Thus, there exists some $c\in \mathbb{Z}_{\left( p\right) }$ such that $b=ac$. Consider this $c$. Now, from $b=ac$, we obtain $v_{p}\left( b\right) =v_{p}\left( ac\right) =v_{p}\left( a\right) +v_{p}\left( c\right) $ (by Proposition 1 (a), applied to $c$ instead of $b$). But $c\in\mathbb{Z}_{\left( p\right) }$ and thus $v_{p}\left( c\right) \geq0$ (by the definition of $\mathbb{Z}_{\left( p\right) }$). Hence, $v_{p}\left( b\right) =v_{p}\left( a\right) +\underbrace{v_{p}\left( c\right) }_{\geq0}\geq v_{p}\left( a\right) $, so that $v_{p}\left( a\right) \leq v_{p}\left( b\right) $. This proves the "$\Longrightarrow$" direction of Proposition 3.
$\Longleftarrow:$ Assume that $v_{p}\left( a\right) \leq v_{p}\left( b\right) $. We must show that $a\mid b$ in $\mathbb{Z}_{\left( p\right) }$.
If $b=0$, then this is obvious (since $b=0=a\cdot0$ in this case). Thus, WLOG assume that $b\neq0$. Hence, $v_{p}\left( b\right) <\infty$, so that $v_{p}\left( a\right) \leq v_{p}\left( b\right) <\infty$ and thus $a\neq 0$. Hence, the rational number $b/a$ is well-defined. Proposition 1 (a) (applied to $b/a$ instead of $b$) yields $v_{p}\left( a\left( b/a\right) \right) =v_{p}\left( a\right) +v_{p}\left( b/a\right) $. Hence, \begin{align} v_{p}\left( b/a\right) =v_{p}\left( \underbrace{a\left( b/a\right) } _{=b}\right) -v_{p}\left( a\right) =v_{p}\left( b\right) -v_{p}\left( a\right) \geq0 \end{align} (since $v_{p}\left( a\right) \leq v_{p}\left( b\right) $). In other words, $b/a\in\mathbb{Z}_{\left( p\right) }$ (by the definition of $\mathbb{Z} _{\left( p\right) }$). Since $b=a\left( b/a\right) $, we thus conclude that there exists some $c\in \mathbb{Z}_{\left( p\right) }$ such that $b=ac$ (namely, $c = b/a$). In other words, $a\mid b$ in $\mathbb{Z}_{\left( p\right) }$. This proves the "$\Longleftarrow$" direction of Proposition 3. Thus, Proposition 3 is fully proven. $\blacksquare$
Proposition 3 characterizes when an element of $\mathbb{Z}_{\left( p\right) }$ is divisible by another. As we have already seen, even if both elements are integers, this is not equivalent to their divisibility in $\mathbb{Z}$. But in the particular case when $a$ is a power of $p$, the relation $a\mid b$ in $\mathbb{Z}_{\left( p\right) }$ is actually equivalent to the analogous relation in $\mathbb{Z}$, as the following proposition shows:
I won't need this proposition in answering your question, so you can skip its proof:
Proof of Proposition 4. Proposition 1 (d) (applied to $n=b$) yields the equivalence $\left( p^{i}\mid b\right) \ \Longleftrightarrow\ \left( v_{p}\left( b\right) \geq i\right) $. Since the statement "$p^{i}\mid b$" is to be understood as "$p^{i}\mid b$ in $\mathbb{Z}$" here, we can rewrite this equivalence as follows: \begin{align} \left( p^{i}\mid b\text{ in }\mathbb{Z}\right) \ \Longleftrightarrow \ \left( v_{p}\left( b\right) \geq i\right) . \end{align}
But $v_{p}\left( p^{i}\right) =i$ (by the definition of $p$-adic valuation). Furthermore, $p^{i}\in\mathbb{Z}\subseteq\mathbb{Z}_{\left( p\right) }$ and $b\in\mathbb{Z}\subseteq\mathbb{Z}_{\left( p\right) }$. Hence, Proposition 3 (applied to $a=p^{i}$) yields that $p^{i}\mid b$ in $\mathbb{Z}_{\left( p\right) }$ if and only if $v_{p}\left( p^{i}\right) \leq v_{p}\left( b\right) $. Hence, we have the following chain of equivalences: \begin{align*} \left( p^{i}\mid b\text{ in }\mathbb{Z}_{\left( p\right) }\right) \ & \Longleftrightarrow\ \left( \underbrace{v_{p}\left( p^{i}\right) }_{=i}\leq v_{p}\left( b\right) \right) \ \Longleftrightarrow\ \left( i\leq v_{p}\left( b\right) \right) \\ & \Longleftrightarrow\ \left( v_{p}\left( b\right) \geq i\right) \ \Longleftrightarrow\ \left( p^{i}\mid b\text{ in }\mathbb{Z}\right) \end{align*} (due to the equivalence $\left( p^{i}\mid b\text{ in }\mathbb{Z}\right) \ \Longleftrightarrow\ \left( v_{p}\left( b\right) \geq i\right) $ that we proved above). This proves Proposition 4. $\blacksquare$
Congruence of two integers modulo a power of $p$ also means the same no matter whether we regard them as elements of $\mathbb{Z}$ or as elements of $\mathbb{Z}_{\left( p\right) }$, as the following corollary shows:
I won't need this corollary in answering your question, so you can skip its proof:
Proof of Corollary 5. The definition of congruence in $\mathbb{Z}_{\left( p\right) }$ shows that we have the following equivalence: \begin{align} \left( a\equiv b\operatorname{mod}p^{i}\text{ in }\mathbb{Z}_{\left( p\right) }\right) \ \Longleftrightarrow\ \left( p^{i}\mid a-b\text{ in }\mathbb{Z}_{\left( p\right) }\right) . \end{align} The definition of congruence in $\mathbb{Z}$ shows that we have the following equivalence: \begin{align} \left( a\equiv b\operatorname{mod}p^{i}\text{ in }\mathbb{Z}\right) \ \Longleftrightarrow\ \left( p^{i}\mid a-b\text{ in }\mathbb{Z}\right) . \end{align} But Proposition 4 (applied to $a-b$ instead of $b$) shows that $p^{i}\mid a-b$ in $\mathbb{Z}_{\left( p\right) }$ if and only if $p^{i}\mid a-b$ in $\mathbb{Z}$. That is, we have the equivalence \begin{align} \left( p^{i}\mid a-b\text{ in }\mathbb{Z}_{\left( p\right) }\right) \ \Longleftrightarrow\ \left( p^{i}\mid a-b\text{ in }\mathbb{Z}\right) . \end{align} Thus, we have the following chain of equivalences: \begin{align*} \left( a\equiv b\operatorname{mod}p^{i}\text{ in }\mathbb{Z}_{\left( p\right) }\right) \ & \Longleftrightarrow\ \left( p^{i}\mid a-b\text{ in }\mathbb{Z}_{\left( p\right) }\right) \ \Longleftrightarrow\ \left( p^{i}\mid a-b\text{ in }\mathbb{Z}\right) \\ & \Longleftrightarrow\ \left( a\equiv b\operatorname{mod}p^{i}\text{ in }\mathbb{Z}\right) \end{align*} (due to the equivalence $\left( a\equiv b\operatorname{mod}p^{i}\text{ in }\mathbb{Z}\right) \ \Longleftrightarrow\ \left( p^{i}\mid a-b\text{ in }\mathbb{Z}\right) $ that we proved above). This proves Corollary 5. $\blacksquare$
Divisibilities and congruences in $\mathbb{Z}_{\left( p\right) }$ satisfy the same basic rules that divisibilities and congruences in $\mathbb{Z}$ (and in any commutative ring) satisfy: For example, divisibility is transitive; congruence modulo any given $c \in \mathbb{Z}_{\left( p\right) }$ is an equivalence relation; congruences modulo one and the same number $c\in\mathbb{Z}_{\left( p\right) }$ can be added, subtracted, multiplied and substituted (in appropriate contexts). All of this is proven in the same way as you would prove the analogous facts for integers.
As a rule, whenever you have a rational number $r$ written as $\dfrac{m}{n}$ with $m,n\in\mathbb{Z}$, and you want to prove that $p\mid m$, you should be thinking of $\mathbb{Z}_{\left( p\right) }$ (and, specifically, of proving that $r\equiv0\operatorname{mod}p$ in $\mathbb{Z}_{\left( p\right) }$). Here are the details:
Proof of Proposition 6. We have $r\equiv0\operatorname{mod}p$ in $\mathbb{Z}_{\left( p\right) }$. In other words, $p\mid r-0$ in $\mathbb{Z}_{\left( p\right) }$. In other words, $p\mid r$ in $\mathbb{Z} _{\left( p\right) }$ (since $r-0=r$).
But $p\in\mathbb{Z}\subseteq\mathbb{Z}_{\left( p\right) }$. Hence, Proposition 3 (applied to $a=p$ and $b=r$) yields that $p\mid r$ in $\mathbb{Z}_{\left( p\right) }$ if and only if $v_{p}\left( p\right) \leq v_{p}\left( r\right) $. Hence, $v_{p}\left( p\right) \leq v_{p}\left( r\right) $ (since $p\mid r$ in $\mathbb{Z}_{\left( p\right) }$). Thus, $v_{p}\left( r\right) \geq v_{p}\left( p\right) =1$ (by the definition of $p$-adic valuation). Also, $v_{p}\left( n\right) \geq0$ (since $n\in\mathbb{Z}$). But $m=nr$ (since $r=\dfrac{m}{n}$), so that $v_{p}\left( m\right) =v_{p}\left( nr\right) =v_{p}\left( n\right) +v_{p}\left( r\right) $ (by Proposition 1 (a), applied to $a=n$ and $b=r$). Hence, $v_{p}\left( m\right) =\underbrace{v_{p}\left( n\right) }_{\geq 0}+\underbrace{v_{p}\left( r\right) }_{\geq1}\geq1$.
Proposition 1 (d) (applied to $1$ and $m$ instead of $i$ and $n$) yields the equivalence $\left( p^{1}\mid m\right) \ \Longleftrightarrow\ \left( v_{p}\left( m\right) \geq1\right) $. Hence, $p^{1}\mid m$ (since $v_{p}\left( m\right) \geq1$). In other words, $p^{1}\mid m$ in $\mathbb{Z}$ (since the divisibility in Proposition 1 (d) is a divisibility in $\mathbb{Z}$). In other words, $p\mid m$ in $\mathbb{Z}$. This proves Proposition 6. $\blacksquare$
Proof of Proposition 7. Clearly, $k\in\mathbb{Z}\subseteq\mathbb{Z}_{\left( p\right) }$ and $\dfrac{1}{k}\in\mathbb{Q}$ (since $k\neq0$). Also, $1 \in \mathbb{Z}$ and thus $v_p\left(1\right) \geq 0$. (Actually, $v_{p}\left( 1\right) =0$, but we don't care.)
Also, from $k\in\left\{ 1,2,\ldots,p-1\right\} $, we obtain $p\nmid k$ in $\mathbb{Z}$. Thus, $v_{p}\left( k\right) =0$ (by the definition of $p$-adic valuation). But the definition of the $p$-adic valuation of rational numbers shows that $v_{p}\left( \dfrac{1}{k}\right) =v_{p}\left( 1\right) -\underbrace{v_{p}\left( k\right) }_{=0}=v_{p}\left( 1\right) \geq0$. Hence, $\dfrac {1}{k}\in\mathbb{Z}_{\left( p\right) }$ (by the definition of $\mathbb{Z} _{\left( p\right) }$). This proves Proposition 7. $\blacksquare$
2. Replacing denominators by binomial coefficients
Next come our two main tricks:
Proof of Lemma 8. We shall work in $\mathbb{Z}_{\left( p\right) }$ throughout this proof (which means that all congruences and divisibilities are understood to be in $\mathbb{Z}_{\left( p\right) }$).
For each $i\in\left\{ 1,2,\ldots,k\right\} $, we have $i\in\left\{ 1,2,\ldots,k\right\} \subseteq\left\{ 1,2,\ldots,p-1\right\} $ (since $k\leq p-1$) and thus $\dfrac{1}{i}\in\mathbb{Z}_{\left( p\right) }$ (by Proposition 7, applied to $i$ instead of $k$). In other words, the $k$ numbers $\dfrac{1}{1},\dfrac{1}{2},\ldots,\dfrac{1}{k}$ belong to $\mathbb{Z}_{\left( p\right) }$. Hence, their product $\dfrac{1}{1}\cdot\dfrac{1}{2}\cdot \cdots\cdot\dfrac{1}{k}$ belongs to $\mathbb{Z}_{\left( p\right) }$ as well (since $\mathbb{Z}_{\left( p\right) }$ is a subring of $\mathbb{Q}$). In other words, $\dfrac{1}{k!}$ belongs to $\mathbb{Z}_{\left( p\right) }$ (since $\dfrac{1}{k!}=\dfrac{1}{1\cdot2\cdot\cdots\cdot k}=\dfrac{1}{1} \cdot\dfrac{1}{2}\cdot\cdots\cdot\dfrac{1}{k}$). In other words, $\dfrac {1}{k!}\in\mathbb{Z}_{\left( p\right) }$.
The definition of $\dbinom{p}{k}$ yields \begin{align*} \dbinom{p}{k} & =\dfrac{p\left( p-1\right) \left(p-2\right) \cdots\left( p-k+1\right) }{k!}=\dfrac{p\cdot\left( p-1\right) \left( p-2\right) \cdots\left( p-k+1\right) }{k!}\\ & =p\cdot\dfrac{\left( p-1\right) \left( p-2\right) \cdots\left( p-k+1\right) }{k!}. \end{align*} Dividing this equality by $p$, we find \begin{align} \dfrac{1}{p}\dbinom{p}{k}=\dfrac{\left( p-1\right) \left( p-2\right) \cdots\left( p-k+1\right) }{k!}=\underbrace{\dfrac{1}{k!}}_{\in \mathbb{Z}_{\left( p\right) }}\underbrace{\left( p-1\right) \left( p-2\right) \cdots\left( p-k+1\right) }_{\in\mathbb{Z}\subseteq \mathbb{Z}_{\left( p\right) }}\in\mathbb{Z}_{\left( p\right) } \end{align} (since $\mathbb{Z}_{\left( p\right) }$ is a subring of $\mathbb{Q}$). Thus, $-\dfrac{1}{p}\dbinom{p}{k}\left( -1\right) ^{k}$ belongs to $\mathbb{Z} _{\left( p\right) }$ as well (since $\mathbb{Z}_{\left( p\right) }$ is a ring). Furthermore, $\dfrac{1}{k}$ belongs to $\mathbb{Z}_{\left( p\right) }$ (by Proposition 7). Hence, we have shown that both sides of the congruence \eqref{darij1.eq.l8.cong} belong to $\mathbb{Z}_{\left( p\right) }$. It remains to actually prove this congruence.
We have $\underbrace{p}_{\equiv0\operatorname{mod}p}-i\equiv -i\operatorname{mod}p$ for each $i\in\left\{ 1,2,\ldots,k-1\right\} $. Multiplying these $k-1$ congruences, we obtain \begin{align*} & \left( p-1\right) \left( p-2\right) \cdots\left( p-\left( k-1\right) \right) \\ & \equiv\left( -1\right) \left( -2\right) \cdots\left( -\left( k-1\right) \right) =\left( -1\right) ^{k-1}\cdot \underbrace{1\cdot2\cdot\cdots\cdot\left( k-1\right) }_{=\left( k-1\right) !}\\ & =\left( -1\right) ^{k-1}\cdot\left( k-1\right) !\operatorname{mod}p. \end{align*} Hence, \begin{align*} \dfrac{1}{p}\dbinom{p}{k} & =\dfrac{1}{k!}\underbrace{\left( p-1\right) \left( p-2\right) \cdots\left( p-k+1\right) }_{\substack{ = \left( p-1\right) \left( p-2\right) \cdots\left( p-\left(k-1\right)\right) \\ \equiv\left( -1\right) ^{k-1}\cdot\left( k-1\right) !\operatorname{mod}p }}\\ & \equiv\dfrac{1}{k!}\cdot\left( -1\right) ^{k-1}\cdot\left( k-1\right) !\qquad\left( \text{since }\dfrac{1}{k!}\in\mathbb{Z}_{\left( p\right) }\right) \\ & =\dfrac{1}{k\cdot\left( k-1\right) !}\cdot\left( -1\right) ^{k-1} \cdot\left( k-1\right) !\qquad\left( \text{since }k!=k\cdot\left( k-1\right) !\right) \\ & =\dfrac{1}{k}\cdot\left( -1\right) ^{k-1}\operatorname{mod}p. \end{align*} Thus, \begin{align} -\underbrace{\dfrac{1}{p}\dbinom{p}{k}}_{\equiv\dfrac{1}{k}\cdot\left( -1\right) ^{k-1}\operatorname{mod}p}\left( -1\right) ^{k}\equiv-\dfrac {1}{k}\cdot\underbrace{\left( -1\right) ^{k-1}\cdot\left( -1\right) ^{k} }_{=\left( -1\right) ^{2k-1}=-1}=-\dfrac{1}{k}\cdot\left( -1\right) =\dfrac{1}{k}\operatorname{mod}p. \end{align} This yields the congruence \eqref{darij1.eq.l8.cong}. Thus, Lemma 8 is proven. $\blacksquare$
3. $S_{a}$ as quasi-Fermat quotients
Let me repeat a definition made in the question (to keep this answer self-contained):
We can now demystify the $S_{a}$ from the question, at least as far as their mod-$p$ congruence class in $\mathbb{Z}_{\left( p\right) }$ is concerned:
Proof of Lemma 9. Let us first work in $\mathbb{Z}$ (so our divisibilities and congruences should be interpreted in $\mathbb{Z}$). Fermat's Little Theorem yields $a^{p}\equiv a\operatorname{mod}p$ and $\left( 1-a\right) ^{p}\equiv1-a\operatorname{mod}p$. Thus, \begin{align} \underbrace{a^{p}}_{\equiv a\operatorname{mod}p}+\underbrace{\left( 1-a\right) ^{p}}_{\equiv1-a\operatorname{mod}p}-1\equiv a+\left( 1-a\right) -1=0\operatorname{mod}p. \end{align} Hence, $p\mid a^{p}+\left( 1-a\right) ^{p}-1$. Thus, $\dfrac{a^{p}+\left( 1-a\right) ^{p}-1}{p}\in\mathbb{Z}$, so that $-\underbrace{\dfrac {a^{p}+\left( 1-a\right) ^{p}-1}{p}}_{\in\mathbb{Z}}\in\mathbb{Z} \subseteq\mathbb{Z}_{\left( p\right) }$.
The definition of $S_{a}$ yields \begin{align} S_{a}=\sum_{k=1}^{p-1}\underbrace{\dfrac{a^{k}}{k}}_{=\dfrac{1}{k}a^{k}} =\sum_{k=1}^{p-1}\underbrace{\dfrac{1}{k}}_{\substack{\in\mathbb{Z}_{\left( p\right) }\\\text{(by Proposition 7)}}}\underbrace{a^{k}}_{\in\mathbb{Z} \subseteq\mathbb{Z}_{\left( p\right) }}\in\mathbb{Z}_{\left( p\right) } \end{align} (since $\mathbb{Z}_{\left( p\right) }$ is a subring of $\mathbb{Q}$). Hence, we have shown that both sides of the congruence \eqref{darij1.eq.l9.cong} belong to $\mathbb{Z}_{\left( p\right) }$. It thus remains to prove this congruence \eqref{darij1.eq.l9.cong}.
We have $1-a=\left( -a\right) +1$, so that \begin{align*} \left( 1-a\right) ^{p} & =\left( \left( -a\right) +1\right) ^{p} =\sum_{k=0}^{p}\dbinom{p}{k}\underbrace{\left( -a\right) ^{k}}_{=\left( -1\right) ^{k}a^{k}}\underbrace{1^{p-k}}_{=1}\qquad\left( \text{by the binomial formula}\right) \\ & =\sum_{k=0}^{p}\dbinom{p}{k}\left( -1\right) ^{k}a^{k}\\ & =\underbrace{\dbinom{p}{0}}_{=1}\underbrace{\left( -1\right) ^{0}} _{=1}\underbrace{a^{0}}_{=1}+\sum_{k=1}^{p-1}\dbinom{p}{k}\left( -1\right) ^{k}a^{k}+\underbrace{\dbinom{p}{p}}_{=1}\underbrace{\left( -1\right) ^{p} }_{\substack{=-1\\\text{(since }p\text{ is odd)}}}a^{p}\\ & \qquad\left( \begin{array} [c]{c} \text{here, we have split off the addends for }k=0\\ \text{and for }k=p\text{ from the sum} \end{array} \right) \\ & =1+\sum_{k=1}^{p-1}\dbinom{p}{k}\left( -1\right) ^{k}a^{k}+\left( -1\right) a^{p}=1+\sum_{k=1}^{p-1}\dbinom{p}{k}\left( -1\right) ^{k} a^{k}-a^{p}. \end{align*} Adding $a^{p}-1$ to both sides of this equality, we find \begin{align} \left( 1-a\right) ^{p}+a^{p}-1 & =1+\sum_{k=1}^{p-1}\dbinom{p}{k}\left( -1\right) ^{k}a^{k}-a^{p}+a^{p}-1\nonumber\\ & =\sum_{k=1}^{p-1}\dbinom{p}{k}\left( -1\right) ^{k}a^{k} . \label{darij1.pf.l9.5} \tag{3} \end{align} Now, in $\mathbb{Z}_{\left( p\right) }$, we have \begin{align*} S_{a} & =\sum_{k=1}^{p-1}\underbrace{\dfrac{1}{k}}_{\substack{\equiv -\dfrac{1}{p}\dbinom{p}{k}\left( -1\right) ^{k}\operatorname{mod} p\\\text{(by \eqref{darij1.eq.l8.cong})}}}a^{k}\equiv\sum_{k=1}^{p-1}\left( -\dfrac{1}{p}\dbinom{p}{k}\left( -1\right) ^{k}\right) a^{k}\\ & =-\dfrac{1}{p}\underbrace{\sum_{k=1}^{p-1}\dbinom{p}{k}\left( -1\right) ^{k}a^{k}}_{\substack{=\left( 1-a\right) ^{p}+a^{p}-1\\\text{(by \eqref{darij1.pf.l9.5})}}}=-\dfrac{1}{p}\left( \left( 1-a\right) ^{p} +a^{p}-1\right) \\ & =-\dfrac{a^{p}+\left( 1-a\right) ^{p}-1}{p}\operatorname{mod} p\qquad\text{in }\mathbb{Z}_{\left( p\right) }. \end{align*} This proves the congruence \eqref{darij1.eq.l9.cong}. Thus, Lemma 9 is proven. $\blacksquare$
We can now solve the question:
Proof of Corollary 10. For each $a\in\mathbb{Z}$, we have $S_{a} \in\mathbb{Z}_{\left( p\right) }$ (indeed, this was proven in our proof of Lemma 9). Thus, in particular, we have $S_{4}\in\mathbb{Z}_{\left( p\right) }$ and $S_{3}\in\mathbb{Z}_{\left( p\right) }$ and $S_{2}\in\mathbb{Z} _{\left( p\right) }$. Hence, $S_{4}+S_{3}-3S_{2}\in\mathbb{Z}_{\left( p\right) }$ (since $\mathbb{Z}_{\left( p\right) }$ is a ring).
But Fermat's Little Theorem yields $2^{p}\equiv2\operatorname{mod}p$ in $\mathbb{Z}$; thus, $p\mid2^{p}-2$ in $\mathbb{Z}$. In other words, there exists an integer $c$ such that $2^{p}-2=pc$. Consider this $c$. Clearly, $c \in \mathbb{Z} \subseteq \mathbb{Z}_{\left( p\right) }$.
(a) We shall work in $\mathbb{Z}_{\left( p\right) }$ throughout this proof (which means that all congruences and divisibilities are understood to be in $\mathbb{Z}_{\left( p\right) }$).
Lemma 9 (applied to $a=4$) yields \begin{align} S_{4}\equiv-\dfrac{4^{p}+\left( 1-4\right) ^{p}-1}{p}=-\dfrac{4^{p}-3^{p} -1}{p}\operatorname{mod}p \end{align} (since $4^{p}+\left( 1-4\right) ^{p}=4^{p}+\underbrace{\left( -3\right) ^{p}}_{\substack{=-3^{p}\\\text{(since }p\text{ is odd)}}}=4^{p}-3^{p}$). Similar computations show \begin{align} S_{3}\equiv-\dfrac{3^{p}-2^{p}-1}{p}\operatorname{mod}p \end{align} and \begin{align} S_{2}\equiv-\dfrac{2^{p}-1^{p}-1}{p}\operatorname{mod}p. \end{align} Combining these three congruences, we obtain \begin{align*} & S_{4}+S_{3}-3S_{2}\\ & \equiv\left( -\dfrac{4^{p}-3^{p}-1}{p}\right) +\left( -\dfrac{3^{p} -2^{p}-1}{p}\right) -3\left( -\dfrac{2^{p}-1^{p}-1}{p}\right) \\ & =\left( -\dfrac{\left( 2^{p}\right) ^{2}-3^{p}-1}{p}\right) +\left( -\dfrac{3^{p}-2^{p}-1}{p}\right) -3\left( -\dfrac{2^{p}-1-1}{p}\right) \\ & \qquad\left( \text{since }4^{p}=\left( 2^{p}\right) ^{2}\text{ and } 1^{p}=1\right) \\ & =\dfrac{-\left( 2^{p}-2\right) ^{2}}{p}\qquad\left( \text{by straightforward computations}\right) \\ & =\dfrac{-\left( pc\right) ^{2}}{p}\qquad\left( \text{since } 2^{p}-2=pc\right) \\ & =-pc^{2}\equiv0\operatorname{mod}p \end{align*} (since $c \in \mathbb{Z} \subseteq \mathbb{Z}_{\left( p\right) }$). This proves Corollary 10 (a).
(b) Corollary 10 (a) yields $S_{4}+S_{3}-3S_{2}\equiv 0\operatorname{mod}p$ in $\mathbb{Z}_{\left( p\right) }$ (which means, in particular, that $S_{4}+S_{3}-3S_{2}\in\mathbb{Z}_{\left( p\right) }$). Hence, Proposition 6 (applied to $r=S_{4}+S_{3}-3S_{2}$) yields $p\mid m$ in $\mathbb{Z}$. This proves Corollary 10 (b). $\blacksquare$