Confused by a proof about harmonic numbers I've been puzzled by a step in D'Aurizio's proof concerning the finiteness of a certain subset $J_p$ of $\mathbf{N}$: $$J_p = \{n : p \text{ divides the numerator of } H_n\}.$$ 
His paper is here: http://arxiv.org/abs/1102.0765. [Note that he refers to $J_p$ unconventionally as $M_p$; everybody, including all the papers he cites, calls this set $J_p$.]
A lemma from within reads
Lemma 2. If $a$, $b$, $c$ are three distinct elements of $J_p$ with $a < b$ and $c - b \equiv 0$ (mod $p$) then $$c - b + a \in J_p.$$
Immediately after the proof of the lemma, the argument becomes hard to follow:

Let us suppose, now, that $J_p$ is infinite. Then at least one residue class in $J_p/(p^2-p)$ is infinite; let $a$ be the smallest positive integer in such a classes. $p^2-p$ belongs to $J_p$, so, by Lemma 2, we have that $$a + (p^2-p)\mathbf{N} \subseteq J_p\tag{1}$$
  holds, and by applying Lemma 2 again, $$(p^2-p)\mathbf{N}\subseteq J_p.\tag{2}$$

Let's pick this apart. An infinite subset of $\mathbf{Z}$ modulo any nonzero integer will split into finitely many classes, so they can't all be finite. Thus, at least one residue class of $J_p$ modulo $p^2-p$ is infinite: pick one such class and call it $S$. Let $a = \min S$. The fact that $p^2 - p$ belongs to $J_p$ is a theorem of Eswarathasan and Levine. So far, so good. 
Questions.


*

*Why is (1) true?

*How does (2) follow from (1)?

 A: I sat down and came up with a thorough debunking.
Write $\mathbf{N} = \{1, 2, 3, \dotsc\}$ and $\mathbf{N}_0 = \mathbf{N} \cup \{0\}$.
First, (2) follows from (1).
Suppose $p > 3$.
We have to show that $(p^2 - p)k \in J_p$ for all $k \ge 1$.
For $k = 1$ this is true by Lemma 5.1 in the founding paper of Eswarathasan and Levine.
For $k > 1$, we apply Lemma 2 to the triplet 
$$(p^2 - p, a + (p^2 - p), a + (p^2 - p)k)$$ 
and get that
$$\big(a + (p^2 - p)k\big) - \big(a + (p^2 - p)\big) + \big(p^2 - p\big) = (p^2 - p)k \in J_p.$$
Next, (1) is almost true. 
Let $S$ be an infinite residue class of $J_p/(p^2 - p)$.
Then $S \subseteq J_p$. 
Let $a = \min S$. 
D'Aurizio's choice in notation suggests that
this number play the role of "$a$" in Lemma 2.
Order and label the elements of $S$ as $a < a_1 < a_2 < \dotsc$. 
Every two members of $S$ are congruent mod $p$ because they are congruent mod $p^2 - p = p(p-1)$.
Thus, every two distinct $a_i$'s are eligible for the roles of "$b$" and "$c$" in Lemma 2.
We have to show that $a + (p^2 - p)k$ is in $J_p$ for all $k \ge 1$. 
Well, $$a_{i+1} - a_i \equiv 0 \pmod{p^2 - p}$$ 
which means the $a_i$'s are separated by jumps 
of size a multiple of $(p^2-p)$.
So there exist positive integers $k_i$ 
such that $$a_{i+1} = a_i + k_i (p^2 - p)$$ 
for all $i \ge 1$.
Choose $k_0 = \min_i k_i$ and let $i_0$ be an index 
at which this minimum is attained. 
Then $k_0 = k_{i_0}$ is (directly proportional to) 
the closest distance between two elements of $S$.
Let $b = a_{i_0}$ and $c = a_{i_0 + 1}$. Then $c - b = (p^2 - p)k_0$.
By Lemma 2 applied to the triplet $(a, b, c)$, we get
$$c - b + a = a + (p^2 - p)k_0 \in J_p;$$ 
by Lemma 2 applied to the triplet $(a + (p^2 - p)k_0, b, c)$, we get 
$$c - b + a + (p^2 - p)k_0 = a + 2(p^2 - p)k_0 \in J_p;$$
by induction it is clear that 
$$a + (p^2 - p)k_0 \mathbf{N} \subseteq J_p. \tag{1'}$$
However, there is no guarantee that $k_0 = 1$. 
The greedy approach above
suggests that (1') is the best we can do;
a "correct" version of (2) is the statement
$$(p^2 - p)(1 + k_0\mathbf{N}_0) \in J_p \tag{2'}$$ 
which follows from (1') and Lemma 2 by choosing the triplet 
$$(p^2 - p, a + (p^2 - p)k_0, a + (p^2 - p)jk_0) \qquad (j > 1).$$
If you look closely, nothing in the proof of (1') used the fact that $S$ is infinite. Roughly speaking, lemma 2 says we can augment any element of $J_p$ by the difference of two others, provided that this difference is divisible by $p$. Therefore, if any two "large" elements of $J_p$ differ by a multiple of $p$ (which is likely), then an inductive argument (analogous to the one preceding (1') above) proves, by way of lemma 2, that $J_p$ is infinite. 
Since we know that $J_p$ is finite for all but three $p < 550$, we begin to suspect lemma 2. 
And indeed, lemma 2 is false. There are plenty of counterexamples.
For one, $J_3 = \{2, 7, 22\}$. The triplet $(2, 7, 22)$ satisfies the hypotheses of Lemma 2 (because $2 < 7$ and $22 - 7 = 15 \equiv 0$ (mod $3$)) but $J_3$ does not contain $22 - 7 + 2 = 17$.
For another, $$J_7 = \{6, 42, 48, 295, 299, 337, 341, 2096, 2390, 14675, 16731, 16735, 102728\}.$$
[Reference for this: the last page of Eswarathasan and Levine and also bottom of page 3 of Keith Conrad's well-written blurb on the problem]. 
Lemma 2 fails for the triplets $(42, 295, 337)$ and $(48, 295, 337)$, and probably many others.
We leave it as an exercise to the reader to find the flaw in D'Aurizio's proof of his Lemma 2.
