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Hello !

I would like prove that the following diophantine equation is unsolvable: $m!+27=n^3$.

Thanks in advance.

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Whe does this question arise? – Yemon Choi May 8 '12 at 5:28
Strictly speaking, I don't think that's a Diophantine equation. . . – Noah Schweber May 8 '12 at 7:43
But the more interesting so... – Felix Goldberg May 8 '12 at 8:28
I think the question is interesting especially in view of GH's answer, and should stay open. – Mark Sapir May 9 '12 at 2:14
There's an excellent paper of Berend and Harmse [Trans. AMS 358 (2005)] which treats equations of the shape $m!=p(x)$ for various polynomials $p(x)$ (generalizing an old problem of Brocard). The (very nice) argument of GH is evident in section 4, applied to, for example, to $p(x)=x (x^2+1)$. – Mike Bennett May 10 '12 at 3:53
up vote 29 down vote accepted

I am happy to report that the equation has no solution. I kept my original response, and put the remaining arguments in the "EDIT" section below.

Here is a quick proof that there are only finitely many solutions.

We use $m!=(n-3)(n^2+3n+9)$. Here $n$ is divisible by $3$, hence $n^2+3n+9$ is not divisible by any prime $p\equiv 2\pmod{3}$. In other words, all the prime divisors $p\equiv 2\pmod{3}$ of $m!$ are contained in $n-3$ with multiplicity. It follows, with the usual notations, that

$$ \frac{\log m!}{3}>\log(n-3)\geq\sum_{p\equiv 2 \ (3), \ p\leq m}v_p(m!)\log p> \sum_{p\equiv 2\ (3), \ p\leq m} \left(\frac{m}{p}-1\right)\log p.$$

The left hand side is $\sim (m\log m)/3$, while the right hand side is $\sim (m\log m)/2$ by Dirichlet's theorem. Hence for large $m$ the inequality must fail.

EDIT. Assume that $m\geq 1000$, and denote by $\chi$ the nontrivial character modulo $3$. Then $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<\sum_{n\leq m}\frac{\chi(n)\Lambda(n)}{n}-\sum_{p\leq m,\ p\neq 3}\frac{\log p}{p^2+p}. $$ This implies, in combination with some ideas of Bordelles (cf. the proof of (4.2) here), that $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+1.53<2.64\ .$$ By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also some classical bounds by Rosser and Schoenfeld (cf. (3.15) and (3.21) here), it follows that $$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-6.1)}{2}\quad\text{for}\quad m>e^{16.5}.$$ Hence $m < e^{16.5}$. I checked with SAGE that in fact $$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.63\quad\text{for}\quad e^{16.5}>m>e^{7}.$$ This can be used to improve the previous bound to $$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-2.92)}{2}\quad\text{for}\quad e^{16.5}>m>e^{7},$$ which in turn forces $m < 1000$. The above shows that all solutions of the original equation satisfy $m < 1000$. However, I checked with SAGE that in this range the equation has no solution.

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Thank you so much GH, now it is interesting find an upper bound for $m$ such that we can begin a computer search. – Roberto Bosch Cabrera May 8 '12 at 14:51
This problem arise (it is the hard part) when we try to solve the equation: $x!+y!+3=Z^3$ which was proposed as $J198$ in Mathematical Reflections. I think that the solution can be found in (1) (2)… but I'm not sure. I'm trying to find an "elementary" solution. – Roberto Bosch Cabrera May 8 '12 at 15:14
It seems that Erdős's paper contains everything needed for a solution. Namely, we are dealing with (8), where $p=3$ and $y=3$. Hence (10) needs to be adjusted slightly: $B_2\leq 27 T(n,6)$. From here it should be easy to finish the solution. At the heart of the argument is (Va), which is a clever explicit substitute of Dirichlet's theorem. – GH from MO May 8 '12 at 16:06
@GH: Thank you for your new Edit proving that $m<10^{12}$. – Roberto Bosch Cabrera May 8 '12 at 23:24

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