How could we solve the following equation (for (m,n)) via a combinatorial method? $1! 3! 5! ... (2n-1)! = m!$ , wherein m and n are positive integers.
Thanks in advance for your help.
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1I guess we may prove that there are a finite number of solutions by comparing the magnitudes of both terms (estimated through Stirling's inequality) and $\nu_2$ of both terms, computed from Legendre's theorem. – Jack D'Aurizio Jan 18 '17 at 23:14
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1By equating $\nu_2$, we get that we must have $m\approx n^2$. But in such a case the magnitude of the LHS does not match the magnitude of the RHS. – Jack D'Aurizio Jan 18 '17 at 23:19
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4Does this answer your question? Find all natural numbers $n > 1$ and $m > 1$ such that $1!3!5!\cdots(2n - 1)! = m!$ – YuiTo Cheng Jul 20 '20 at 08:52
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The only solutions are given by $(n,m)\in\{(1,1),(2,3),(3,6),(4,10)\}$.
If we define $$ \nu_2(a) = \max\{h\in\mathbb{N}: 2^h\mid a\} $$ and assume that $n$ is large enough, by Legendre's theorem we get that $m$ has to be close to $n^2$ to ensure $\nu_2(1!3!\cdots(2n-1)!)=\nu_2(m!)$. Since $n^2>2(2n-1)$, there is some prime in the interval $[2n,m]$ that is a divisor or $m!$ but is not a divisor of $1!3!\cdots(2n-1)!$.
We may check the cases up to $n=12$ by hand, then the last argument proves the claim.
Jack D'Aurizio
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Thank you for your answer. I am trying to solve the equation via a combinatorial approach rather than an algebraic one. If we consider m objects in which, $1$ of them is of first type, $3$ of them of second type, ..., and $(2n-1)$ of them of "n"th type, we will have $\frac{m!}{1!3!...(2n-1)!}$ ways to arrange these $m$ objects in a line. Obviously, $m$ would be equal to $m = 1+3+...+(2n-1)+ q = n^2+q$ in which $q$ is the remaining distinct objects. However, I don't know how can I continue to solve the problem using this approach ... – Shahram Khazaie Jan 18 '17 at 23:37
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@ShahramKhazaie: I am not aware of combinatorial arguments for proving that a sort-of diophantine equation only has a finite amount of solutions. In this case, however, we may exploit arithmetic and analytic arguments (like in many irrationality proof) through $\leq, \nu_2$ and $\nu_p$. – Jack D'Aurizio Jan 18 '17 at 23:42
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$(2n-1)!(2n-3)!(2n-5)!>(4n)!$ for $n > t_0$ (simple to prove this by induction), where k is the minimal number
where are at least one prime number between $2n$ and $4n$
So, $n < t_0$
Just precalc this $t_0$
kotomord
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