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The Brocard Problem shows three factorials that can be expressed as $n!+1=m^2$ or equivalently $n!=k(k+2)$. So any time a single factorial can be put in the form of $k(k+2)$, it will belong to Brocard Problem series. However the series has only three elements and extensive search has not uncovered new elements.

It turned out that it is much easier to generate sums of factorials that satisfy $(n! + i! +\ldots j!)=m^2-1$.
Here are some simple examples:

$4! + 4!= 48 = 6*(6+2)= 7^2 -1$
$5! + 5! + 5!=360= 18*(18+2)= 19^2 -1$
$5! + 6!= 840= 28*(28+2)= 29^2 - 1$
$4! + 4! + 5!= 168= 12*(12+2) = 13^2 -1$
$4! + 4! + 5! + 5!= 288 = 16*(16+2) = 17^2 -1$
$4! + 5! + 6! + 7! + 8!= 214*(214+2) =215^2 -1$

More can be generated, for example, we have:

$8! + 2! + 3! + 3*4!= 200*(200+2)= 201^2 -1$
$6! + 2! + 3!=26*(26+2)=27^2-1$

We start with expressing $10!$ as a difference of squares then we convert the smaller square into factorials:

$10!=1905^2-15^2$
$15^2=225=5! +4*4! +3! +2! +1$
so we get:
$10! +5! +4*4! +3! +2! =1905^2-1$

To generate these kind of relations involving sums of factorials, we start with a number $n=k(k+2)$ then we try to fit factorials that sum up to the desire number $m^2-1$. I suppose there is more than one way to express a given number as a sum of factorials especially for large numbers.

I think the reason we can't generate more Brocard factorials ( single factorials of the form $n!=m^2-1$) is because the pattern is broken. Already we have a "discontinuity" of the pattern when going from $5!$ to $7!$ since $6!$ is not a Brocard factorial. Numbers of the form $k*(k+2)$ ( both even and odd ) can be found on the third diagonal of the multiplication table ( the diagonal starting with $3$). The number $6$ is already not of that form since $6!=6*10*12$. The simple multiplication by 6 destroyed the pattern $10*12=10*(10+2)=5!$. Larger number (single factorials) than $7!$ will not exhibit the pattern $k*(k+2)$ since we keep adding factors of the form $j*(j+i)$ with $i\gt2$. I am not saying it's a proof. I am just pointing out a pattern.

Question 0: Why is it much easier to generate sums of factorials that satisfy Brocard relation $n! + i! +\ldots+ j! =m^2 -1$

Question 1: Are there large number factorials that cannot be put as a sum of factorials satisfying $n! +i!+\ldots+ j!=m^2-1$?

Question 2: Will it be easier to settle Brocard problem if, instead of using the multiplications to evaluate $n!$, the factorials are expressed as polynomials with higher and higher degrees and open the door to using knowledge of the mathematics of polynomials?

Note: The quadratic equation $n!=k*(k+2)$ can be transform into the quartic $n!=m^4+2m^3-m^2-2m$. Solving the quartic for a given $n!$ will instantly tell if a single factorial is a Brocard's one just like the quadratic does.

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  • $\begingroup$ "$n! = k *(k+2)$ can be transform into the quartic $n! = m^4 + 2 m^3 - m^2 - 2 m$"? Why should $k$ be of the form $m(m-1)$?? $\endgroup$ Commented May 30, 2023 at 15:40
  • $\begingroup$ in fact the numbers involved in $k∗(k+2)$ are $j(j+1)$, twice a triangular number $(2,6,12,20...)$ and the one aligned with is of the form $i(i+3)$ forming the diagonal $(4,10,18,28,...)$. It's in fact $k∗(k+2)$ should read $i∗(i+3)∗j∗(j+1)$. First, it's easier to deal with an equation with only one variable instead of $2$ but also because I noticed (by looking at the multiplication table) that $4!=4∗6$, $5!=10∗12$...so I started using $k∗(k+2)$. To get a one variable quadratic or quartic, $i∗(i+3)$ was converted to $(j−1)(j−1+3)$ so we get the quartic $(m−1)∗(m+2)∗m∗(m+1)$. . $\endgroup$ Commented May 30, 2023 at 16:20
  • $\begingroup$ (cont); to make things simpler, I am not using $n!=n*(n-1)*(n-2)*...1$. I am using half the multiplications needed as shown in the post below so that $4!=4*6$. scicomp.stackexchange.com/questions/42510/… $\endgroup$ Commented May 30, 2023 at 16:27
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    $\begingroup$ Every natural number can be expressed as a sum of factorials $k!$ with coefficients in $\{0,1,\ldots,k-1\}$. See factorial base. $\endgroup$ Commented May 30, 2023 at 19:07
  • $\begingroup$ Sorry, that should be $\{0,1,\ldots,k\}$. $\endgroup$ Commented May 30, 2023 at 21:42

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Simply put, sums of a few factorials are a lot more common than factorials themselves, so it's likely that a lot more of them will have a particular form.

There are, for example, $1023$ natural numbers $< 11!$ that are sums of distinct factorials (not counting $0!$ and $1!$ as different). $12$ of them are of the form $m^2-1$:
$$ \eqalign{3 &= 1! + 2! = 2^2 - 1\cr 8 &= 2! + 3! = 3^2 - 1\cr 24 &= 4! = 5^2 - 1\cr 120 &= 5! = 11^2 - 1\cr 728 &= 2! + 3! + 6! = 27^2 - 1\cr 840 &= 5! + 6! = 29^2 - 1\cr 5040 &= 7! = 71^2 - 1\cr 45368 &= 2! + 3! + 7! + 8! = 213^2 - 1\cr 46224 &= 4! + 5! + 6! + 7! + 8! = 215^2 - 1\cr 363608 &= 2! + 3! + 6! + 9! = 603^2 - 1\cr 403224 &= 4! + 8! + 9! = 635^2 - 1\cr 3674888 &= 2! + 3! + 6! + 7! + 8! + 10! = 1917^2 - 1\cr }$$

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  • $\begingroup$ I can add this one to the list: $10! + 12! = 21969^2 - 81^2$. Now we just express $81^2=6561= 7! +2*6! + 3*4! + 3! + 2! + 1$. So $10! + 12! + 7! + 2*6! + 3*4! + 3! + 2!=21969^2 - 1$ $\endgroup$ Commented May 30, 2023 at 19:53
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    $\begingroup$ That doesn't go in my list because of the $2 * 6!$ and $3 * 4!$: my list is for sums of distinct factorials. But there is $ 2! + 3! + 7! + 8! + 9! + 10! + 11! + 12! + 13! + 14! + 15! = 1183893^2 - 1$. $\endgroup$ Commented May 30, 2023 at 21:47

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