Let us say that three consecutive positive integers $(m-1,m,m+1)$ have a big prime factor if the largest prime factor $p$ of $N:=(m-1)m(m+1)$ satisfies $e^p>N$.


I ckecked that it is true for all $m<10^8$, except for $m=3,9,15,49,55,99,351,441,2431$ where $p=3,5,7,7,11,11,13,17,19$, respectively.


**Question**: Is it true for all $m>2431$?


*Application*: A positive answer would solve a problem in Section D25 of Richard Guy's [*Unsolved Problems in Number Theory*](https://doi.org/10.1007/978-0-387-26677-0).



> 
> Simmons notes that $n! = (m - 1)m(m + 1)$ for $(m, n) = (2, 3)$, $(3, 4)$, $(5, 5)$, and $(9, 6)$, and asks if there are other solutions. More generally he asks if there are any other solutions of $n! + x = x^k$. This is a variation on the question of asking for $n!$ to be the product of $k$ consecutive integers in a nontrivial way ($k \ne n + 1 - j!$). Compare **B23**.
> 
> 
> 


Note that $n! > e^n$ iff $n \ge 6$. But I checked for $m \le 2431$ that $n! = (m-1)m(m+1)$ iff $$(m,n) = (2,3), (3,4), (5,5), (9,6).$$ So for $m>2431$, if $e^p>N$ and $N = n!$, then $p! \ge n!$, but $p<n$ because $p$ is a prime factor of $n!$ (and $n>2$), contradiction.




---


Our question for three consecutive positive integers can be generalized to $k \ge 3$ consecutive positive integers, where the bound $e^p$ could be optimized. Now for $k=2$ it is not clear that $e^p$ (or even $p!$) works because I found large exceptions.



```
sage: m=123200
sage: factor(m*(m+1))
2^6 * 3^6 * 5^2 * 7 * 11 * 13^2
sage: factorial(13)>m*(m+1)
False


sage: m=2697695
sage: factor(m*(m+1))
2^5 * 3^2 * 5 * 7^3 * 11^2 * 13 * 17 * 19 * 29
sage: ceil(exp(29))>m*(m+1)
False

```

On the contrary, for $k=2$ we can wonder whether, for a given prime $p>2$, there are infinitely many $m$ such that the biggest prime factor of $m(m+1)$ is $\le p$. The case $p=3$ reduces to the existence of infinitely many couple of non-negative integers $(a,b)$ with $\lvert 2^a-3^b\rvert = 1$, but it is false (see why at [distance between powers of 2 and powers of 3](https://mathoverflow.net/q/116840/34538)). Now what if $p$ is large *enough*? In other words:


**Bonus question**: Is there an integer $n$ such that the set of integers $m(m+1)$ whose prime factors are less than $n$, is infinite?



Do consecutive integers have a big prime factor?
Thanks to Szpiro’s/$abc$ conjecture, the answer to your question is true for sufficiently large $N$, up to a factor of $3$ times the primes $p$ (which factor, I believe, can be removed if one is a bit careful in the estimation of the prime factors of $N$ below $3\log m$ in the proof below). Note that the problem you wish to solve with the answer to your question can still be solved (for sufficiently large numbers) if you weaken your hypothesis to $e^{kp}>N$ for some $k\ge1$ (indeed, $p\gg\log N$ suffices for your problem as $\log n!\gg n$). Here, I’d simply resort to $k=3$ (actually, any $k>9/4=2.25$ suffices) to allow for a simpler proof.


For a rational elliptic curve $E$ with minimal discriminant $\Delta\_E$ and conductor $f\_E$, let
$$\sigma\_E:=\frac{\log|\Delta\_E|}{\log f\_E}\,.$$ Szpiro’s conjecture states that for every $\varepsilon>0$, there are only finitely many elliptic curves with $\sigma\_E\ge 6+\varepsilon$.


Now, consider the rational elliptic curve
$$E\colon y^2=x(x-1)(x-m^2)$$
for a given natural number $m$. By well-known/standard results, the above is a minimal model and so the minimal discriminant is $$\Delta\_E=(4m^2(m^2-1))^2\,.$$
Note that, in your notation, $N=m^3-m$, and so $\Delta\_E=(4mN)^2$. In particular, $E$ has multiplicative reduction at all odd prime factors of $\Delta\_E$ but additive reduction at $2$; thus, the conductor $f\_E$ is $$f\_E=2^{n}\prod\_{2<p|N}p\,,$$ for some $n\in\{2,3,4,5,6\}.$ Now, working asymptotically (we write $\sim$ for *asymptotically equal to* and $\gtrsim$ for *asymptotically greater than or equal to*), then assuming to the contrary that $e^{kp}\le N$ for $k=3$, or better still some $k>9/4$, we obtain
\begin{align}
\log f\_E&= n\log 2 +\sum\_{2<p|N}\log p\\
&\le n\log 2+ \sum\_{p\le\frac{1}{k}\log N}\log p\\
&\sim n\log 2+ \sum\_{p\le\frac{3}{k}\log m}\log p\\
&\sim n\log 2+\frac{3}{k}\log m\,,
\end{align}
where we have used the prime number theorem $\sum\_{p\le x}\log p\sim x$ in the final step. This implies that
\begin{align}
\sigma\_E=\frac{\log|\Delta\_E|}{\log f\_E}&=\frac{\log 16m^4(m^2-1)^2}{\log f\_E}\\
&\sim\frac{\log 16m^8}{\log f\_E}\\
&\gtrsim\frac{4\log2+8\log m}{n\log 2+ \frac{3}{k}\log m}\\
&\sim \frac{8}{3}k>6,
\end{align}
which contradicts Szpiro’s conjecture for sufficiently large $N$.


