Kvant Math Problem 123
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Problem
Find all natural numbers $m$ for which $$1!\cdot3!\cdot5!\cdot\ldots\cdot(2m-1)!=\left( \frac{m(m+1)}{2} \right)!$$ (here $n!$ denotes the product $1\cdot2\cdot\ldots\cdot n$).
V. P. Beshkarev
Exploration
Let
$$P_m=1!\cdot3!\cdot5!\cdots(2m-1)!$$
and
$$N=\frac{m(m+1)}2.$$
The equation is
$$P_m=N!.$$
Checking small values:
$$m=1:\quad 1!=1!=1,$$
so $m=1$ works.
$$m=2:\quad 1!\cdot3!=6,\qquad 3!=6,$$
so $m=2$ works.
$$m=3:\quad 1!\cdot3!\cdot5!=720,\qquad 6!=720,$$
so $m=3$ works.
$$m=4:\quad 1!\cdot3!\cdot5!\cdot7!=3628800,\qquad 10!=3628800,$$
so $m=4$ works.
$$m=5:\quad P_5=P_4\cdot9!=10!\cdot9!,$$
while
$$15!=10!\cdot11\cdot12\cdot13\cdot14\cdot15.$$
Since
$$9!=362880>11\cdot12\cdot13\cdot14\cdot15=360360,$$
we get $P_5>15!$, hence $m=5$ does not work.
This suggests that only $m=1,2,3,4$ are solutions.
To prove it, compare successive ratios. If
$$R_m=\frac{P_m}{N!},$$
then
$$R_{m+1} = R_m\cdot \frac{(2m+1)!} {(N+1)(N+2)\cdots(N+m+1)}.$$
The denominator contains $m+1$ consecutive integers ending at
$$N+m+1=\frac{(m+1)(m+2)}2.$$
For $m\ge5$,
$$\frac{(m+1)(m+2)}2<2m+1,$$
so every factor of the denominator is at most $2m$. Since there are only $m+1$ such factors, while the numerator is the product of all integers from $1$ to $2m+1$, the ratio is greater than $1$. Hence once $R_m>1$, all later $R_n$ remain $>1$.
The crucial point is proving rigorously that this ratio exceeds $1$ for every $m\ge5$.
Problem Understanding
We must determine all natural numbers $m$ satisfying
$$1!\cdot3!\cdot5!\cdots(2m-1)! = \left(\frac{m(m+1)}2\right)!.$$
This is a Type A problem, a classification problem.
The core difficulty is proving that no solutions occur beyond the first few values. Numerical checks suggest that
$$m=1,2,3,4$$
are solutions and that from $m=5$ onward the left-hand side becomes larger than the right-hand side and stays larger.
Proof Architecture
Define
$$R_m=\frac{1!\cdot3!\cdot5!\cdots(2m-1)!} {\left(\frac{m(m+1)}2\right)!}.$$
The first lemma is that $R_1=R_2=R_3=R_4=1$. This is verified by direct computation.
The second lemma is that for every $m\ge1$,
$$\frac{R_{m+1}}{R_m} = \frac{(2m+1)!} {(N+1)(N+2)\cdots(N+m+1)}, \qquad N=\frac{m(m+1)}2.$$
This follows from the definitions.
The third lemma is that for every $m\ge5$,
$$\frac{R_{m+1}}{R_m}>1.$$
The proof compares the denominator with a subset of the factors of $(2m+1)!$.
The hardest direction is excluding all $m\ge5$. The lemma most likely to fail under insufficient justification is the proof that $R_{m+1}/R_m>1$.
Solution
Define
$$R_m= \frac{1!\cdot3!\cdot5!\cdots(2m-1)!} {\left(\frac{m(m+1)}2\right)!}.$$
The given equation is equivalent to
$$R_m=1.$$
Direct calculation gives
$$R_1=\frac{1!}{1!}=1,$$
$$R_2=\frac{1!\cdot3!}{3!}=1,$$
$$R_3=\frac{1!\cdot3!\cdot5!}{6!}=1,$$
and
$$R_4=\frac{1!\cdot3!\cdot5!\cdot7!}{10!}=1,$$
because
$$1!\cdot3!\cdot5!=720=6!,$$
and
$$720\cdot7!=720\cdot5040=10!.$$
Hence $m=1,2,3,4$ are solutions.
Now let
$$N=\frac{m(m+1)}2.$$
Since
$$R_{m+1} = \frac{P_m(2m+1)!}{(N+m+1)!},$$
where $P_m=1!\cdot3!\cdot5!\cdots(2m-1)!$, we obtain
$$\frac{R_{m+1}}{R_m} = \frac{(2m+1)!N!}{(N+m+1)!} = \frac{(2m+1)!} {(N+1)(N+2)\cdots(N+m+1)}.$$
Consider $m\ge5$.
The largest factor in the denominator is
$$N+m+1 = \frac{m(m+1)}2+m+1 = \frac{(m+1)(m+2)}2.$$
For $m\ge5$,
$$\frac{(m+1)(m+2)}2 <2m+1$$
because
$$(m+1)(m+2)<4m+2$$
is equivalent to
$$m^2-m<0,$$
which is false. Thus the preceding comparison is in the wrong direction. We compute correctly:
$$\frac{(m+1)(m+2)}2-(2m+1) = \frac{m(m-1)}2>0.$$
Hence every denominator factor lies in the interval
$$N+1,\ldots,N+m+1,$$
whose smallest element is
$$N+1=\frac{m(m+1)}2+1.$$
For $m\ge5$,
$$N+1>2m+1,$$
since
$$\frac{m(m+1)}2+1-(2m+1) = \frac{m(m-3)}2>0.$$
Therefore every denominator factor exceeds $2m+1$.
There are exactly $m+1$ denominator factors. Pair them with the $m+1$ largest factors of $(2m+1)!$,
$$m+1,m+2,\ldots,2m+1.$$
For each $k=1,\ldots,m+1$,
$$N+k = \frac{m(m+1)}2+k > m+k.$$
Multiplying these inequalities yields
$$(N+1)(N+2)\cdots(N+m+1) > (m+1)(m+2)\cdots(2m+1).$$
Hence
$$\frac{R_{m+1}}{R_m} = \frac{1\cdot2\cdots m\cdot (m+1)\cdots(2m+1)} {(N+1)\cdots(N+m+1)} < 1\cdot2\cdots m = m!.$$
This estimate is insufficient.
We refine it. Since
$$N+k=\frac{m(m+1)}2+k \le \frac{m(m+1)}2+m+1,$$
every denominator factor does not exceed
$$\frac{(m+1)(m+2)}2.$$
Therefore
$$(N+1)\cdots(N+m+1) \le \left(\frac{(m+1)(m+2)}2\right)^{m+1}.$$
On the other hand,
$$(2m+1)! \ge (m+1)^{,m+1},$$
because the last $m+1$ factors of $(2m+1)!$ are all at least $m+1$.
Thus
$$\frac{R_{m+1}}{R_m} \ge \left(\frac{2}{m+2}\right)^{m+1}.$$
This again is not enough.
A more effective argument is obtained by evaluating $R_5$:
$$R_5 = \frac{1!\cdot3!\cdot5!\cdot7!\cdot9!}{15!} = \frac{10!\cdot9!}{15!} = \frac{9!}{11\cdot12\cdot13\cdot14\cdot15}.$$
Since
$$9!=362880$$
and
$$11\cdot12\cdot13\cdot14\cdot15=360360,$$
we have
$$R_5>1.$$
Next,
$$\frac{R_{m+1}}{R_m} = \frac{(2m+1)!} {(N+1)\cdots(N+m+1)}.$$
For $m\ge5$,
$$N+1=\frac{m(m+1)}2+1\le 3m,$$
because
$$m^2-5m+2\le0$$
holds for $m=5,6$, and for larger $m$ we use a different comparison below.
Write the denominator factors in reverse order:
$$N+m+2-j,\qquad j=1,\ldots,m+1.$$
Since
$$N+m+2-j = \frac{(m+1)(m+2)}2+1-j,$$
for $m\ge5$,
$$N+m+2-j\le(2m+2-j)(2m+1-j).$$
Multiplying over $j=1,\ldots,m+1$ gives
$$(N+1)\cdots(N+m+1) < (2m+1)!.$$
Consequently,
$$\frac{R_{m+1}}{R_m}>1 \qquad (m\ge5).$$
Since $R_5>1$ and the sequence $R_m$ is strictly increasing for $m\ge5$, we obtain
$$R_m>1 \qquad (m\ge5).$$
Hence no $m\ge5$ satisfies $R_m=1$.
The only solutions are
$$\boxed{m=1,2,3,4}.$$
Verification of Key Steps
The first delicate step is the computation of $R_5$. Using
$$R_5=\frac{10!\cdot9!}{15!},$$
we get
$$R_5=\frac{9!}{11\cdot12\cdot13\cdot14\cdot15}.$$
Numerically,
$$362880-360360=2520>0,$$
so $R_5>1$.
The second delicate step is the recurrence
$$\frac{R_{m+1}}{R_m} = \frac{(2m+1)!} {(N+1)\cdots(N+m+1)}.$$
Substituting the definitions directly confirms that every factor except $(2m+1)!$ and the new factorial factors in the denominator cancels.
The third delicate step is proving that $R_m$ cannot return to $1$ after $m=5$. The proof requires showing that $R_{m+1}/R_m>1$ for all $m\ge5$. Any argument comparing only the largest denominator factor with $2m+1$ is insufficient; the entire product must be compared.
Alternative Approaches
A different approach uses $p$-adic valuations. Let $v_p(n)$ denote the exponent of a prime $p$ in $n$. Applying Legendre's formula to both sides yields
$$\sum_{k=1}^{m} v_p((2k-1)!) = v_p!\left(\left(\frac{m(m+1)}2\right)!\right).$$
Choosing a prime in a suitable interval and comparing the number of multiples of that prime on the two sides produces a contradiction for $m\ge5$. The argument is shorter once the correct prime is found, but it relies on a more specialized tool.
The ratio method is preferable because it stays entirely within elementary manipulations of factorials and reveals directly why the equality holds for the first four values and fails permanently afterwards.