<s>[INST] 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?
 [/INST] 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.
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