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 $1$, I have $2$ issues (for better searchability for future reader, I asked in single post).
(i) Please read the following paragraph writtten on page 293 –
First observe that by the well-known theorem of Sylvester and Schur, there is always a prime greater than >$k$ which divides $(n + 1)… (n + k)$, since $n > k$. Such a prime divides only one of the $k$ factors,
so $n + k > (k + 1)^l$, whence $n > k^l \cdots (2)$.
If equation $(1)$ is true I understand $n + k > (k + 1)^l$ since there is a prime greater than $> k$ and that prime must have power $l$. But how do we derive $$n > k^l?$$
I mean, $n + k > (k + 1)^l$ is independent of $n > k^l$, then how this two inequalities are related?
(ii) On page 294 it is written that –
$$ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= \frac{kl^2}{2n-kl}$$
If geometric series is used then it should be $ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= l \cdot \frac{1}{1-\frac{kl}{2n}} = l \cdot \frac{2n}{2n-kl}$ (by multiplying $\frac{2n}{2n}$), not $\frac{kl^2}{2n-kl}$.
So questions are: why $n > k^l$ and $ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= \frac{kl^2}{2n-kl}$?
Best Answer
(i) is evident by an application of binomial theorem. Noting that $l > 2$:
$$n+k > (k+1)^l =\sum_{j=0}^l \binom ljk^j > \binom ll k^l+\binom l1 k^1=k^l+lk$$
Hence $n > k^l+(l-1)k > k^l$.
Now for (ii), it is indeed by sum of geometric series.
However the first term is not $1$, but $\dfrac {kl}{2n}$. This gives
$$ l\sum_{i=2}^{\infty} (\frac{kl}{2n})^{i-1}= l \cdot \frac{\frac{kl}{2n}}{1-\frac{kl}{2n}} = l \cdot \frac{kl}{2n-kl}=\frac{kl^2}{2n-kl}$$
as expected.