Do consecutive integers have a big prime factor?

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.

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 $$p because $$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). 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?

• The answer to your bonus question is no by Stormer's theorem (en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem) - for any 'smoothness threshold' there are only finitely many consecutive pairs of smooth integers. (As that wikipedia page notes, Thue-Siegel-Roth is enough to just say no, but Stormer's theorem gives much more explicit information.) Jul 27, 2022 at 16:44
• Shorey and Tijdeman, together and separately, had several papers in the 1970s about the largest prime factor of various strings of integers. It might be worth looking through their papers, also those of Cam Stewart. Jul 27, 2022 at 23:38
• Solutions to $n!=(m-1)m(m+1)$ would imply values 1 or 2 in oeis.org/A355192 Jul 29, 2022 at 14:02

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 $$f_E=2^{n}\prod_{2 for some $$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$$, we obtain \begin{align} \log f_E&= n\log 2 +\sum_{2 where we have used the prime number theorem $$\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$$.
Shorey and Tijdeman, On the greatest prime factors of polynomials at integer points, Compositio Mathematica, tome 33, no 2 (1976), p. 187-195 http://www.numdam.org/item?id=CM_1976__33_2_187_0 note that if $$f$$ is a polynomial with integer coefficients and at least two distinct roots, and we write $$P(m)$$ for the greatest prime factor of $$m$$, then $$P(f(n))\gg\log\log n$$ where the implied constant depends only on the polynomial $$f$$. They say the degree $$3$$ case follows from work of Keates. In fact, they prove something a little stronger, but not nearly strong enough to get the $$p>3\log n$$ that OP asks for.