Solve factorial equal to product of other factorials, $a! = b! \times 5! \times 3!$

Question

General equations of factorials equal to the product of other factorials, e.g. $a! = b! \times c! \times d!$, have been asked before on this site and turn out to be an open problem, though only four non-trivial solutions to such equations are known.*

The very specific question $a! = 7! \times 5! \times 3!$ has also been asked here before. Despite being a straightforward question, in the sense that it is purely computational, it received answers using a range of different methods, and is still somewhat "interesting" because it relates to one of the few non-trivial solutions to the general case.

So, I would like to ask for all non-negative integer solutions to $a! = b! \times 5! \times 3!$. This seems a good "intermediate" question between the previous two: sufficiently more specific than the general case that I believe it can be solved, for example using methods that generalise the problem $a!=7!5!3!$, but rather more interesting than that specific case since it isn't purely computational and will include additional (trivial) solutions, as well as the non-trivial one! I will post an attempt at an answer, but the previous question suggests there will be a wide range of possible approaches so I would be very interested to see how other people tackle it.

(I'm hoping that seeing some methods for solving a more general problem like $a!=b!5!3!$ might at least hint at why finding non-trivial questions to such equations is so rare. A couple of other "intermediate" questions suggest themselves: $a!=7!b!c!$, $a!=5!b!c!$ and $10!=b!c!d!$ would be harder in the sense of solving for three variables, but each would catch two of the non-trivial solutions, while $a!=7!b!c!d!$ would catch three of the four. These might make good follow-up questions, unless someone posts an answer here that easily extends to handle all these cases, but aside from these it seems there are few other "factorial is a product of factorials" problems that would be suited for Math SE without being either too trivial or too difficult.)

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$(*)$ According to this answer on MathOverflow, these will be the only non-trivial solutions if Baker's explicit abc conjecture is correct (see Nair and Shorey's 2016 paper in the Journal of Number Theory).

Answer 1

Your equation can be written as:

$$a! = 720 \times b!$$

Let's enumerate all the ways to write $720$ as the product of consecutive integers:

• $720$
• $10 \times 9 \times 8$
• $6 \times 5 \times 4 \times 3 \times 2$
• $6 \times 5 \times 4 \times 3 \times 2 \times 1$

It turns out that no factor of 720 satisfies the equations $n(n+1) = 720$ or $n(n+1)(n+2)(n+3) = 720$, so there are no expansions with two or four factors. Nor can there be an expansion with 7 or more distinct factors, because the product would have to be at least $7! > 6! = 720$. But there are expansions with 1, 3, 5, and 6 factors.

We want both sides to be a factorial, so just set $b$ to be 1 less than the smallest factor of each expansion.

• $720! = 720 \times 719!$
• $10! = 10 \times 9 \times 8 \times 7!$
• $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1!$
• $6! \times 5 \times 4 \times 3 \times 2 \times 1 \times 0!$

This gives us the solution set:

$$(a, b) \in \{ (720, 719), (10, 7), (6, 1), (6, 0) \}$$

Answer 2

Generalising an answer to $a!=7!5!3!$, consider the order of magnitude of the left and right sides. Factorials can be approximated well using Stirling's approximation but even crude bounds help. Factorials grow very quickly, so (at least past, say, $3!=6$) two factorials are either "exactly the same" or "very far apart". If $a!=b!5!3!$ then the right-hand side will be dominated by the largest factorial, $\max\{b!,5!\}$, but $b!5!3!$ won't be "very much" bigger than $\max \{b!,5! \}$ so it seems $a$ cannot be "much" larger than $\max \{ b,5 \}$. (In fact no non-trivial solutions are known for any equation $a!=b!\times c! \times \cdots$ where, without loss of generality, $b \ge c \ge \cdots$, for which $a > b + 3$.)

Let's write $a=b+k$ and evaluate the known factorials so $(b+k)!=720b!$; clearly $k \ge 1$ and our hope (justified by handwaving above) is that $k$ can't be much larger than one, so we should have only a few cases to check. We need to solve $$(b+1)(b+2)\cdots (b+k) = 720 $$ so an obvious inequality is $$ (b+1)^k \le 720 \le (b+k)^k $$

and if $k>1$ these inequalities become strict. It would be useful to dispose of $k=1$ for this reason: when $k=1$ the above collapses to $b+1=720$ so we obtain $b=719$, $a=b+k=720$, and $720!=719!5!3!$ is a solution.

We could also make a better estimate using the arithmetic mean of $b+1, b+2, \dots, b+k$ which is $b + (k+1)/2$, so that $(b + (k+1)/2)^k \approx 720$. The AM-GM inequality then gives that $(b + (k+1)/2)^k \ge (b+1)\cdots (b+k)$, and since $k>1$ we can make this inequality strict. We now have an improved upper bound, and taking the $k$-th root obtain

$$ b+1 < \sqrt[k]{720} < b + \frac{k+1}{2} \implies \sqrt[k]{720} - \frac{k+1}{2} < b < \sqrt[k]{720} - 1$$

For each $k$ we can see which values of $b$ are plausible, then check if any give $(b+1)\cdots (b+k)=720$.

For $k=2$, $\sqrt{720} = 26.83\dots$ gives $25.33\dots < b < 25.83\dots$ with no integer solution.

For $k=3$, $\sqrt[3]{720} = 8.96\dots$ gives $6.96\dots < b < 7.96\dots$ and hence $b = 7$, $a=b+k=10$. Indeed $8 \times 9 \times 10 = 720$, so $10! = 7! \times 5! \times 3!$ is a solution.

For $k=4$, $\sqrt[4]{720} = 5.18\dots$ gives $2.68\dots < b < 4.18\dots$ so $b=3$ or $b=4$. But neither work, since $4 \times 5 \times 6 \times 7 > 720$ and so clearly $5 \times 6 \times 7 \times 8 > 720$ also.

For $k=5$, $\sqrt[5]{720} = 3.72\dots$ gives $0.72\dots < b < 2.72\dots$ so $b=1$ or $b=2$. For $b = 1$, we need to check $2 \times 3 \times 4 \times 5 \times 6 = 720$, which it does. So we have found the solution $6! = 1! \times 5! \times 3!$. For $b = 2$, clearly the product $3 \times 4 \times 5 \times 6 \times 7 > 720$.

For $k=6$, $\sqrt[6]{720} = 2.99\dots$ gives $-0.51 \dots < b < 1.99 \dots$ and moreover for $k>6$, this means $\sqrt[k]{720} < 3$ and $b < 2$. So it only remains to check $b=0$ and $b=1$ for $k \ge 6$. Either way, $b! = 1$ so our original equation becomes $a!=720$. Since $a=b+k \ge 6$, we check if $6!=720$. It does, so $6! = 0! \times 5! \times 3!$ is a solution. But we can't have a solution for $b=1, k\ge 6$ or $b=0, k\ge 7$ or else $a!=(b+k)!\ge 7!>720$.

Overall we obtain the exact solutions $720! = 719! \times 5! \times 3!$, $10! = 7! \times 5! \times 3!$, $6! = 1! \times 5! \times 3!$ and $6! = 0! \times 5! \times 3!$. Examining our results for $k=2$ and $k=4$ more closely, our method has also found the "near(ish) misses":

$$\frac{27!}{25!5!3!} = 0.975; \frac{28!}{26!5!3!} = 1.05; \frac{6!}{5!3!2!} = 0.5; \frac{7!}{5!3!3!} = 1.1\dot{6} $$

Note that using the AM-GM inequality for $(b+1)(b+2)\cdots (b+k)$ ended up giving us an improved lower bound for $b$, but this didn't turn out to be essential. In the end the upper bound $b < \sqrt[k]{720} - 1$ was more important, since $\log_3 720 = 5.98\dots < 6$ (i.e. $720 < 729 = 3^6$) meant that for $k \ge 6$ we have $b<2$ so $b \in \{0,1\}$. This is why our procedure terminates at the sixth step: once we checked products of at least six consecutive integers starting at $(b+1)\in \{1,2\}$ we knew we had found all solutions. It's clear this procedure will work for all questions of the form $a! = b! \times n$ where $n$ is a given integer, and take $\lceil \log_3{n} \rceil$ steps (where $\lceil \dots \rceil$ is the ceiling function i.e. round the log up). Had we persisted with our cruder lower bound $\sqrt[k]{720} - k < b$, we'd have terminated at the same stage but needed to check a few more cases manually along the way (e.g. for $k=2$ we'd have had $24.83\dots < b < 25.83\dots$ and needed to see if $b=25$ works), which, while inconvenient, would not be fatal to our endeavour.

For $a! = b! \times n$ we see we need to check one case when $k=1$ ($a=n$, $b=n-1$, which always works); for $2 \le k \le \lfloor \log_3{n} \rfloor$ and using the stricter bounds, $b$ lies in the interval $(\sqrt[k]{n} - (k+1)/2, \sqrt[k]{n} - 1)$ which is open with width $(k-1)/2$ so contains at most $\lfloor k/2 \rfloor$ integer values to check; for $k = \lceil \log_3{n} \rceil$ we need to check two cases ($b=0$ and $b=1$, although their details will be identical). So in the worst case the number of possible solutions this procedure needs to check is

$$1 + (1 + 1) + (2 + 2) + \dots + (\lfloor \frac{\lfloor \log_3{n} \rfloor}{2} \rfloor + \lfloor \frac{\lfloor \log_3{n} \rfloor}{2} \rfloor) + 2$$

which is $\mathcal{O}\left((\log_3{n})^2\right)$.