Find all natural numbers $n > 1$ and $m > 1$ such that $1!3!5!\cdots(2n - 1)! = m!$
I have been thinking about coming up with some inequalities which would narrow the possible range of pairs $(n, m)$, however the best I have been able to find so far are $m \ge 2n - 1$ and $m \lt n^2$, which is clearly not enough.
Update.
The question has tag combinatorics as it is from a book about combinatorics, so there must be at least partly combinatorial solution.
On
Note that every prime divisor of $1!3!5!\cdots(2n-1)!$ occurs at least twice, except for perhaps $2n-1$ (if it's prime). By Bertrands Postulate, we already know that $m!$ has one prime divisor that occurs only once. (See Can n! be a perfect square when n is an integer greater than 1?) A slightly stronger version of the theorem will give that there are at least two primes in $m!$ with multiplicity $1$. Indeed, Wikipedia says that for $x\geq25$ there are at least two primes between $x$ and $(1+0.2)^2x=1.44x$. Thus there are no solutions for $m\geq50$ (and you might even reduce this upper bound by checking some cases by hand - that is, looking for $m$ such that there are at least two prime divisors of $m!$ with multiplicity one).
The only solutions are $(2,3),(3,6),(4,10)$.
On
Idea: show that $1!3!5!\cdots(2n-1)!\geqslant(4n-1)!$, so that $m\geq4n-1$. By Bertrands Postulate it will follow that there are no solutions because there is a prime between $2n-1$ and $m$.
It suffices that $1!3!5!\cdots(2n-3)!\geqslant2n(2n+1)\cdots(4n-1)$. Let $n\geq12$. We have
$$\begin{array}{c}(2n-3)!&=1\cdot&\cdots&n\cdot&n+1&\cdots&\cdot(2n-5)&\cdot(2n-4)&\cdot(2n-3)\\ &\geq&2^{n-2}&\cdot\frac{2n}2&\cdot\frac{2n+2}2&\cdots&\cdot\frac{4n-10}2&\cdot\frac{4n-8}2&\cdot\frac{4n-6}2\\ (2n-5)!&=1\cdot&\cdots&n\cdot&n+1&\cdots&\cdot(2n-5)\\ &\geq&2^{n-4}&&\cdot\frac{2n+1}2&\cdots&\cdot\frac{4n-11}2\\ \end{array}$$ so $(2n-5)!(2n-3)!\geq2n(2n+1)\cdots(4n-10)\cdot(4n-8)(4n-6)$. There are $7$ factors missing to get $2n(2n+1)\cdots(4n-1)$. We'll get these factors from the remaining factorials.
Since $x!\geq2\cdot3x\geq2(x+10)$ for $x\geq5$, we have $$\begin{array}{l}&(2n-19)!&(2n-17)!&(2n-15)!&(2n-13)!&(2n-11)!&(2n-9)!&(2n-7)!\\ \geq&(4n-9)&\cdot(4n-7)&\cdot(4n-5)&\cdot(4n-4)&\cdot(4n-3)&\cdot(4n-2)&\cdot(4n-1)\end{array}$$ (note that $2n-19\geq5$ because $n\geq12$) so $(2n-19)!\cdots(2n-3)!\geq2n(2n+1)\cdots(4n-1)$.
Checking $n<12$:
for $m>1$, $m!$ has a prime divisor with multiplicity $1$ so it suffices to check primes $2n-1$:
Let $$N:=m! =1!\>3!\>5!\cdots(2n-1)!$$ for certain numbers $m$, $n\in{\mathbb N}_{\geq1}$, and denote by $p$ the exponent of $2$ in the prime decomposition of $N$. Then one has on the one hand $$p=\left\lfloor{m\over2}\right\rfloor+\left\lfloor{m\over4}\right\rfloor+\ldots <m$$ and on the other hand $$p\geq\sum_{k=1}^n\left\lfloor{2k-1\over2}\right\rfloor={(n-1)n\over 2}\ .$$ This implies $$m>{(n-1)n\over2}\ .$$ On the other hand, by Bertrand's postulate, we must have $m<2(2n-1)$, because otherwise $m!$ would contain a prime factor not present in $(2n-1)!$. Now $${(n-1)n\over2}<2(2n-1)$$ enforces $n\leq8$. But we can do better, since for small values of $2n-1$ there are many more primes available than Bertrand's postulate guarantees. We therefore set up the following table: $$\matrix{n&&1&2&3&4&5&6&7&8\cr 2n-1&&1&3&5&7&9&11&13&15\cr m>&&0&2&3&6&10&15&21&28\cr}$$ This table show that already for $n\geq5$ any "admissible" $m!$ would contain a prime factor $>2n-1$. So it remains to check the cases $n\in[4]$, which lead to $$N\in\{1, \>6, \>720, \>3628800\}=\{1!,\>3!,\>6!,\>10!\}\ .$$ It follows that there are exactly $4$ pairs $(m,n)$ of the required kind.