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

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).

• You might want to clarify what is $a_{i_j}$ since that is not obvious from the context that you gave. Nov 7, 2020 at 16:14
• @CalvinLin I have edited the post plz see. Nov 7, 2020 at 16:24
• I see there's a bounty for "a canonical answer". Is Calvin's answer insufficiently "canonical"? Wouldn't it be better to engage with Calvin, to make it clear what troubles you about Calvin's answer? Nov 15, 2020 at 22:39

(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 $$p^\alpha \mid p^{\sum k_i} \mid (k-m)!(m-1)!$$.

• @ConsiderNon-TrivialCases Be explicit with what parts you are stuck with. IE Write it out so that I can understand what you need explaining. Nov 7, 2020 at 17:03
• Thanks for your help but don't get it yet, if it is not too much trouble to elaborate last two line: " 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 follow.".., showing the derivation by expanding through indices it might be helpful, also don't see how $p^{k_i} + 1 \mid n+m$. Nov 7, 2020 at 17:07
• A) That's by definition. Multiply all the $p^{k_i} || i-m$ statements together, what do you get? Ditto for the second. B) Likewise $K_i \leq k_i$ so sum that up and we are done. C) No, that's not what I said/meant. I'm considering cases. Let me edit. Nov 7, 2020 at 17:07
• By "write it out", I meant write out your working and thought process to demonstrate where you're at, not just "quote whatever I said and throw it all back at me". Nov 7, 2020 at 17:11

$$i-m$$ 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 $$(k-1)!$$.