Product of factorials divided by factorial to produce perfect square Let $ S = 1! ~2!~\dotsm ~100! $. Prove that there exists a unique positive integer $k$ such that $S/k!$ is a perfect square. 
I thought this was a cute, fun problem and I did solve it, but any alternative methods that you guys would use?
 A: More generally:

Theorem 1. Let $n$ be a positive integer divisible by $4$ such that neither $n/2$ nor $n/2+1$ is a perfect square. Let $S = 1! \cdot 2! \cdot \cdots \cdot n!$. There exists a unique positive integer $k$ such that $S / k!$ is a perfect square. This integer $k$ is $n/2$.

The two ingredients that go into the proof of this are the following two lemmas:

Lemma 2. Let $n$ be a nonnegative integer. Then,
  \begin{align}
\dfrac{1! \cdot 2! \cdot \cdots \cdot \left(2n\right)!}{n!} = 2^n \cdot \left(\prod_{i=1}^n \left(\left(2i-1\right)!\right)\right)^2 .
\end{align}

Lemma 2 is Exercise 3.5 (c) in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019, and anyway is easy to prove (e.g., by induction on $n$ if you don't want to match factors).

Lemma 3. Let $u$ and $v$ be two distinct positive integers such that $u!/v!$ is the square of a rational number. Then, we have $\left| u-v \right| = 1$, and the number $\max\left\{u,v\right\}$ is a perfect square.

Proof of Lemma 3. We know that $u!/v!$ is the square of a rational number. In other words, $u! / v! = r^2$ for some rational $r$. Hence, $v! / u! = 1 / \underbrace{\left(u! / v!\right)}_{= r^2} = 1 / r^2 = \left(1/r\right)^2$. Thus, $v! / u!$ is the square of a rational number, too. Hence, our situation is symmetric in $u$ and $v$. Thus, we can WLOG assume that $u \geq v$ (since otherwise, we can swap $u$ with $v$). Assume this.
Hence, $u! / v! = u \left(u-1\right) \cdots \left(v+1\right)$. Thus, $u! / v!$ is an integer. In other words, $r^2$ is an integer (since $u! / v! = r^2$). Thus, $r$ is a square root of an integer, and therefore is either an integer or irrational. Since $r$ is not irrational (because $r$ is rational), we thus conclude that $r$ is an integer. Thus, $r^2$ is a perfect square. In other words, $u! / v!$ is a perfect square (since $u! / v! = r^2$).
We have $u \geq v$, so that $u-v \geq 0$ and thus $\left| u-v \right| = u-v$. Also, $\max\left\{u,v\right\} = u$ (since $u \geq v$).
Let us now prove that $\left| u-v \right| \leq 1$. Indeed, assume the contrary. Thus, $\left| u-v \right| > 1$. Hence, $u-v = \left| u-v \right| > 1$, so that $u > v+1$. Thus, $u \geq v+2$ (since $u$ and $v+1$ are integers). But $u! / v! = u \left(u-1\right) \cdots \left(v+1\right)$. Thus, $u! / v!$ is a product of at least $2$ consecutive positive integers (the "at least $2$" part here comes from the fact that $u \geq v+2$). But a result of Erdos (P. Erdos, Note on products of consecutive integers, J. London Math. Soc. 14 (1939), pp. 194--198) shows that any product of at least $2$ consecutive positive integers is never a perfect square. Hence, we conclude that $u! / v!$ is not a perfect square. This contradicts the fact that $u! / v!$ is a perfect square. This contradiction shows that our assumption was false. Hence, $\left| u-v \right| \leq 1$ is proven.
We have $\left|u-v\right| \neq 0$ (since $u$ and $v$ are distinct). Combining this with $\left| u-v \right| \leq 1$, we obtain $\left| u-v \right| = 1$.
It remains to show that the number $\max\left\{u,v\right\}$ is a perfect square.
Indeed, $u-v = \left|u-v\right| = 1$, and thus $u = v+1$. Hence, $u! / v! = \left(v+1\right)! / v! = v+1 = u$, so that $u = u! / v! = r^2$. Thus, $u$ is a perfect square (since $r$ is an integer). In other words, $\max\left\{u,v\right\}$ is a perfect square (since $\max\left\{u,v\right\} = u$). This completes the proof of Lemma 3. $\blacksquare$
Proof of Theorem 1. We know that $n/4$ is a positive integer (since $n$ is a positive integer divisible by $4$). Hence, $n/2$ is a positive integer as well (since $n/2 = 2 \left(n/4\right)$). Hence, Lemma 2 (applied to $n/2$ instead of $n$) yields
\begin{align}
\dfrac{1! \cdot 2! \cdot \cdots \cdot \left(2\left(n/2\right)\right)!}{\left(n/2\right)!} = 2^{n/2} \cdot \left(\prod_{i=1}^{n/2} \left(\left(2i-1\right)!\right)\right)^2 = \left(2^{n/4} \prod_{i=1}^{n/2} \left(\left(2i-1\right)!\right)\right)^2 .
\end{align}
In view of $1! \cdot 2! \cdot \cdots \cdot \left(2\left(n/2\right)\right)! = 1! \cdot 2! \cdot \cdots \cdot n! = S$, this rewrites as
\begin{align}
\dfrac{S}{\left(n/2\right)!} = \left(2^{n/4} \prod_{i=1}^{n/2} \left(\left(2i-1\right)!\right)\right)^2 .
\end{align}
The number $2^{n/4} \prod\limits_{i=1}^{n/2} \left(\left(2i-1\right)!\right)$ on the right hand side of this equality is an integer (since $n/4$ is an integer). Thus, the right hand side of this equality is a perfect square. Therefore, the left hand side is a perfect square as well. In other words, $\dfrac{S}{\left(n/2\right)!}$ is a perfect square. Hence, there exists at least one positive integer $k$ such that $S / k!$ is a perfect square, namely $k = n/2$.
It remains to prove that there exists at most one such integer. In other words, it remains to prove that if $w$ is any positive integer $k$ such that $S/k!$ is a perfect square, then $w = n/2$.
We shall first prove a weaker claim:
(1) If $u$ and $v$ are two distinct positive integers $k$ such that $S/k!$ is a perfect square, then $\max\left\{u,v\right\}$ is a perfect square and we have $\left| u-v \right| = 1$.
[Proof of (1): Let $u$ and $v$ be two distinct positive integers $k$ such that $S/k!$ is a perfect square.
We know that $u$ is a positive integer $k$ such that $S/k!$ is a perfect square.
In other words, $u$ is a positive integer such that $S/u!$ is a perfect square.
Hence, there exists some integer $x$ such that $S / u! = x^2$.
Similarly, there exists some integer $y$ such that $S / v! = y^2$.
Consider these $x$ and $y$.
Note that $S = 1! \cdot 2! \cdot \cdots \cdot n! \neq 0$ (since $1!, 2!, \ldots, n!$ are nonzero). Thus, $S / u! \neq 0$, so that $x^2 = S / u! \neq 0$ and thus $x \neq 0$. Hence, $y/x$ is a well-defined rational number (since $x$ and $y$ are integers). This number satisfies $\left(y/x\right)^2 = \underbrace{y^2}_{=S/v!} / \underbrace{x^2}_{=S/u!} = \left(S/v!\right) / \left(S/u!\right) = u!/v!$.
Hence, $u!/v!$ is the square of a rational number (namely, of $y/x$). Thus, Lemma 3 shows that we have $\left| u-v \right| = 1$, and that the number $\max\left\{u,v\right\}$ is a perfect square. This proves (1).]
Now, let $w$ be a positive integer $k$ such that $S/k!$ is a perfect square. We shall prove that $w = n/2$.
Indeed, assume the contrary. Thus, $w \neq n/2$. Hence, $w$ and $n/2$ are distinct. But $w$ and $n/2$ are two positive integers $k$ such that $S/k!$ is a perfect square. Hence, (1) (applied to $u = w$ and $k = n/2$) yields that $\max\left\{w,n/2\right\}$ is a perfect square and that we have $\left| w-n/2 \right| = 1$. We have $\max\left\{w,n/2\right\} \neq n/2$ (since $\max\left\{w,n/2\right\}$ is a perfect square but $n/2$ is not). If we had $w \leq n/2$, then we would have $\max\left\{w,n/2\right\} = n/2$, which would contradict $\max\left\{w,n/2\right\} \neq n/2$. Hence, we cannot have $w \leq n/2$. Thus, we have $w > n/2$. Hence, $\left| w-n/2 \right| = w-n/2$, so that $w-n/2 = \left| w-n/2 \right| = 1$ and thus $w = n/2 + 1$. But $w > n/2$, so that $\max\left\{w,n/2\right\} = w = n/2 + 1$. Thus, $n/2 + 1$ is a perfect square (since $\max\left\{w,n/2\right\}$ is a perfect square). This contradicts the fact that $n/2 + 1$ is not a perfect square.
This contradiction completes our proof of $w = n/2$.
Forget that we fixed $w$. We thus have shown that if $w$ is any positive integer $k$ such that $S/k!$ is a perfect square, then $w = n/2$. As we know, this completes our proof of Theorem 1. $\blacksquare$
A: Other answers here have given general results, but it seems worthwhile to give a self-contained answer specific to the problem with $S=1!2!3!4!\cdots99!100!$, assuming the OP wants $S/k!$ to be a perfect integer square.  
As observed in the other answers, 
$$S=1!(1!\cdot2\cdot1)3!(3!\cdot2\cdot2)\cdots99!(99!\cdot2\cdot50)=(1!)^2(3!)^2\cdots(99!)^22^{50}\cdot50!$$
shows that $S/50!=(1!3!99!2^{25})^2$ is a perfect square. It remains that show that $k=50$ is the unique value such that $S/k!$ is a perfect integer square.
Let's assume that $S/k!$ is a perfect integer square. Then $50!/k!=(S/k!)/(S/50!)$ is a rational square. Now, as Will Jagy notes, it's clear we need $k\lt101$, since otherwise $S/k!$ has the prime $101$ in its denominator (and hence is not an integer). But now it's clear we need $47\le k\lt53$:  If $k\lt47$, $50!/k!$ will have a single $47$ in its numerator, while if $53\le k\lt101$, it will have a single $53$ in its denominator.  And the remaining five cases (other than $k=50$) are easily ruled out:
$$\begin{align}
50!/47!&=50\cdot49\cdot48=2^5\cdot3\cdot5^2\cdot7^2\\
50!/48!&=50\cdot49=2\cdot5^2\cdot7^2\\
50!/49!&=50=2\cdot5^2\\
51!/50!&=51=3\cdot17\\
52!/50!&=52\cdot51=2^2\cdot3\cdot13\cdot17
\end{align}$$
Remark: If we drop the assumption that the OP wants $S/k!$ to be an integer square, it's still possible to prove that $k=50$ is the only choice, but the simplest proof I can think of invokes Bertrand's Postulate to limit $k\le100$. I'd be pleased to see a proof that doesn't make use of the guaranteed existence of a prime between $k/2$ and $k$.
A: We can rewrite $S$ as
\begin{align*}
S &= 1^{100} \cdot 2^{99} \cdot 3^{98} \cdot 4^{97} \dotsm 99^2 \cdot 100 \\
&= 1^{100} \cdot 2 \cdot 2^{98} \cdot 3^{98} \cdot 4 \cdot 4^{96} \dotsm 99^2 \cdot 100 \\
&= 1^{100} \cdot 2^{98} \cdot 3^{98} \cdot 4^{96} \dotsm 99^2 \times 2 \cdot 4 \dotsm 100.
\end{align*}Clearly, the expression to the left of the times sign is a perfect square. Let $X$ denote the expression to the right of the times sign. We have
[X = 2 \cdot 4 \dotsm 100 = \left( 2 \cdot 1 \right) \left( 2 \cdot 2 \right) \dotsm \left( 2 \cdot 50 \right) = 2^{50} \cdot 50!.]Thus, $S/50!$ is a perfect square. Now, we will prove that $50$ is the only such positive integer.
Assume for the sake of contradiction that $S/a!$ is a perfect square for some positive integer $a \neq 50$. Then we have
$$$S = m^2 \cdot a! = n^2 \cdot 50! \quad \quad (*)$$for some nonnegative integers $m$ and $n$. If $a \ge 101$, then $101 \mid a!$ and $101 \mid S.$ But from the definition of $S,$ clearly $101 \not{\mid} S$ so we have a contradiction. Thus, $a \le 100$. We now have two cases:
Case 1: $0 < a < 50$. From $(*)$, we have
$$\left( \frac{m}{n} \right)^2 = (a + 1)(a + 2) \dotsm 50,$$so $(a + 1)(a + 2) \dotsm 50$ is a perfect square. If $a + 1 \le 47$, then $(a + 1)(a + 2) \dotsm 50$ will contain only one factor of $47$ (a prime) and will never be a perfect square. We can now easily verify that $48 \cdot 49 \cdot 50 = 2^5 \cdot 3 \cdot 5^2 \cdot 7^2$, $49 \cdot 50 = 2 \cdot 5^2 \cdot 7^2$, and $50 = 2 \cdot 5^2$ are not perfect squares.
Case 2: $50 < a \le 100$. Similar to the first case, we have
$$\left( \frac{n}{m} \right)^2 = 51 \cdot 52 \dotsm a,$$so $51 \cdot 52 \dotsm a$ is a perfect square. If $a \ge 53$, then $51 \cdot 52 \dotsm a$ will contain only one factor of $53$ (a prime) and will never be a perfect square. We can once again easily check that $51 \cdot 52 = 2^2 \cdot 3 \cdot 13 \cdot 17$ and $51 = 3 \cdot 17$ are not perfect squares.
Thus, $50$ is the only positive integer $k$ such that $S/k!$ is a perfect square.
