The Product of Consecutive Integers is Never a Power: Lemma 2 (Research Paper Study)

Question

P. Erdos and J. L. Selfridge proved in the paper THE PRODUCT OF CONSECUTIVE INTEGERS IS NEVER A POWER (click here), that the equation $(n + 1) \cdots(n + k)=x^l \cdots (1)$ has no solution in integers with $k > 2, l > 2, n > 0$. There is a lemma $2$ on page 294, 295 -

LEMMA 2. By deleting a suitably chosen subset of $\pi(k-1)$ of the numbers $a_i (1 \leq i \leq k)$, we have $$a_{i_l}...a_{i_k'},|(k -1)!\cdots (9)$$ where $k’=k -\pi(k-1)$.

For each prime $p < k-1 $ we omit an $a_m$ for which $ n + m$ is divisible by $p$ to the highest power. If $1 \leq i \leq k$ and $i \neq m$, the power of $p$ dividing $n +i$ is the same as the power of $p$ dividing $i-m$. Thus $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)!$.

I understand $p | i-m$ but how does $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)! \:\:$ ?

Edit:

Here, $n + i= a_ix^l$, where $a_i$ is $l^{th}$-power free and all its prime factors are less than $k$ (supposing Theorem 2 is false, see page 293 of the paper (click here) for the definition).

Answer 1

(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Just write it out.

Let $ p^{k_i} || n+i$.

For $ i \neq m$, we now show that $p^{k_i} || i-m$. By definition of $m$, $p^{k_i} \mid n+m$.

• Case 1: $ p^{k_i + 1} \mid n+m$. Verify that we do indeed have $ p^{k_i} || (n+i)-(n+m) = i-m$.
• Case 2: $ p^{k_i} || n+m$. Then $p^{ k_i} \mid (n+i)-(n+m)$ and $ |i-m| \leq p^{k_i + 1}$ hence $ p^{ k_i} || i-m$.

Let $K=k_i \pmod{l}$ (in reduced modulo class). Then by definition, $ p^{K_i} || a_i$, and $K_i \leq k_i$.

Hence, we have $ p^{\sum k_i } || \prod_{i\neq m}(i-m) = (k-m)!(m-1)! $ and $p^{\sum K_i } || \prod a_i $.

Since $ \sum K_i \leq \sum k_i$, the desired result follows.
Namely, with $\alpha = \sum K_i$, we have that $ p ^\alpha || \prod a_i$ and $ \alpha \leq \sum k_i$ so $ p^\alpha \mid p^{\sum k_i} \mid (k-m)!(m-1)!$.

Answer 2

$i-m$ is divisible by $p$, just like $(i-m)$, $p$ divides $(m-i)$, from $(n+m)-(n+i) \equiv 0 \pmod p \implies m-i \equiv 0 \pmod p$, look at the definition in pdf file.

See that $(i-m)$ , $(m-i)$ act as a a factor of $(k-m)!(m-1)!$.

it is easy to find that $(k-m)!(m-1)!| (k-1)!$.

That is why $p^{\alpha}$ divides $a_{i_{1}} \cdots a_{i_{k'}}$, gives $p^{\alpha}$ divides $(k-m)!(m-1)!$and , $p^{\alpha}$ divides $(k-1)!$.