Smallest n needing 6 Numbers less than n so that Product of Factorials is Square
Theorem
Let $n \in \Z_{>0}$ be a positive integer.
Then it is possible to choose at most $6$ positive integers less than $n$ such that the product of their factorials is square.
The smallest $n$ that actually requires $6$ numbers to be chosen is $527$.
Proof
Obviously the product cannot be a square if $n$ is a prime.
For $n$ composite, we can write:
- $n = a b$
where $a, b \in \Z_{>1}$.
Then:
| \(\ds \) | \(\) | \(\ds n! \paren {n - 1}! \paren {a!} \paren {a - 1}! \paren {b!} \paren {b - 1}!\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n a b \paren {\paren {n - 1}! \paren {a - 1}! \paren {b - 1}!}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {n! \paren {a - 1}! \paren {b - 1}!}^2\) |
which is a square.
Hence no more than $6$ factorials is required.
To show that $527$ is the smallest that actually requires $6$, observe that:
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It might be worth extracting some of the below statements into lemmata, for example: "If $n$ is itself square, then so is $n! \paren {n - 1}!$" and "... Then $n! \paren {n - 1}! b! \paren {b - 1}!$ is square" -- they're really easy to prove, even I can do them :-) but it takes more than a glance to recognise that they are true.
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If $n$ is itself square, then so is $n! \paren {n - 1}!$.
If $n$ is not square-free, write $n = a^2 b$, where $b$ is square-free.
Then $n! \paren {n - 1}! b! \paren {b - 1}!$ is square.
If $n$ is divisible by $2$, write $n = 2 m$.
Then $\paren {2 m}! \paren {2 m - 1}! \paren {m!} \paren {m - 1}! \paren {2!}$ is square.
If $n$ is divisible by $3$, write $n = 3 m$.
Then $\paren {3 m}! \paren {3 m - 1}! \paren {2 m}! \paren {2 m - 1}! \paren {3!}$ is square.
If $n$ is divisible by $5$, write $n = 5 m$.
Then $\paren {5 m}! \paren {5 m - 1}! \paren {m!} \paren {m - 1}! \paren {6!}$ is square.
If $n$ is divisible by $7$, write $n = 7 m$.
Then $\paren {7 m}! \paren {7 m - 1}! \paren {5 m}! \paren {5 m - 1}! \paren {7!}$ is square.
If $n$ is divisible by $11$, write $n = 11 m$.
Then $\paren {11 m}! \paren {11 m - 1}! \paren {7 m}! \paren {7 m - 1}! \paren {11!}$ is square.
The remaining numbers less than $527$ that are not of the above forms are:
- $221, 247, 299, 323, 377, 391, 403, 437, 481, 493$
Each of the following is a product of $5$ factorials which is square:
- $221! \, 220! \, 18! \, 11! \, 7!$
- $247! \, 246! \, 187! \, 186! \, 20!$
- $299! \, 298! \, 27! \, 22!$
- $323! \, 322! \, 20! \, 14! \, 6!$
- $377! \, 376! \, 29! \, 23! \, 10!$
- $391! \, 389! \, 24! \, 21! \, 17!$
- $403! \, 402! \, 33! \, 30! \, 14!$
- $437! \, 436! \, 51! \, 49! \, 28!$
- $481! \, 479! \, 38! \, 33! \, 22!$
- $493! \, 491! \, 205! \, 202! \, 7!$
The fact that $527$ has no such representation can be verified by a direct (but lengthy) computation.
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Sources
- 1976: P. Erdos and R.L. Graham: On Products of Factorials (Bulletin of the Institute of Mathematics Academia Sinica Vol. 4, no. 2)
- 1983: François Le Lionnais and Jean Brette: Les Nombres Remarquables: $527$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $527$
- 1994: Richard K. Guy: Unsolved Problems in Number Theory (2nd ed.): $\mathbf {B 23}$: Equal products of factorials.