Summing of factorials to produce perfect cubes I was playing around with factorials the other day, and I realized that $4!+5!$ is a perfect square. Perplexed by this result, I started looking for other pairs of factorials that produce a perfect square when added together (unbeknownst to me, I had stumbled across a well-known open problem in number theory). I could only find three: $1!+4!$, $1!+5!$, and $1!+7!$. Since $4!+5!$ seemed like an outlier, I decided to focus on the three which satisfy the formula $n!+1=m^2$. I turned to the main Math.SE chatroom, asking if there were any more, which is when I was informed that this is called Brocard's problem. Since it's a well-known open problem, I decided to put my own spin on it. It has probably been done before, but I can't find anything online.What about the equation $n!+1=m^3$? As far as I know, there is no pair of integers that satisfies this equation. What if the conditions are relaxed? This brings me to the following question I came up with yesterday: How many perfect cubes can be expressed as the sum of two or more unique integer factorials without subtractions, divisions, or multiplications of factorials?Notice I said unique, meaning any factorial cannot appear more than once in a sum. With that in mind, I've only found three:$$ \begin{align}&2!+3!=8=2^3\\&1!+2!+4!=27=3^3\;\;\;\Bbb{and}\\&1!+2!+3!+6!=729=9^3\,.\end{align}$$Are there more solutions, or are these the only three?Using this link, I've concluded any additional solution must have the factorials add up to be greater than $341^3$. Since I hardly have any experience with programming, I can't go much further on this problem... Any and all help would be greatly appreciated.Note: My question is NOT a duplicate of this one. I'm not restricting my problem to the partial sums $S_n$ of the series $\displaystyle \sum_{k=1}^nk!$.
 A: This is just a comment, but notice that if $a_{1}
 < a_{2} < \ldots < a_{n}$, then $a_{1}! + a_{2}! + \ldots + a_{m}!$  is divisible by $a_{1}!$, and if $a_{2} > a_{1} + 1$ if $a_{1}$ is even or $a_{2} > a_{1}$ if $a_{1}$ is odd, then each $a_{j}!$ is divisible by a higher power of $2$ than $a_{1}!$  is for each $j > 1.$ In either of these cases, the highest power of $2$ dividing $a_{1}!$ is an integer multiple of $3$ ( though possibly zero, which happens only when $a_{1} = 1$). This puts constraints on $a_{1}$  when $a_{2} > a_{1}+1.$ It's possible to argue similarly for other primes $p \leq a_{1}$ when there is an integer multiple of $p$ between $a_{1}$ and $a_{2}$ (including $a_{2}$ itself)- where we obtain that the sum of the digits in the $p$-adic expansion of $a_{1}$ is a multiple of $p-1.$ However, I can't see this approach alone being sufficient to answer the question ( it is no help when $a_{1} = 1$, for example. On the other hand, when $a_{1} = 2,$ it forces $a_{2} = 3$, and if we try to find a longer example with $a_{1}= 2$ and $a_{2} = 3$ we find it impossible).
