nicetry007 wrote: For $ p = 2$, it is trivially true. Let $ p \;\geq\; 3$. $ Q(p) = \prod_{i = 1}^{p - 1}k^{2k - p - 1} = \left( \frac {\prod_{i = 1}^{p - 1}k^{k}}{\left( (p - 1)! \right)^{(p + 1)/2}}\right)^{2}$ It is enough to show that the term inside the parenthesis is an integer. Let $ q$ be a prime less than $ p$. The exponent of $ q$ in the denominator is given by $ \frac {(p + 1)}{2}\left (\lfloor\frac {p - 1}{q}\rfloor + \lfloor\frac {p - 1}{q^{2}}\rfloor + \cdots + \lfloor\frac {p - 1}{q^{t}}\rfloor \right)$ In the numerator, the terms that are multiples of $ q$ are as follows: $ q^{q}, (2q)^{2q}, \cdots, \left( \lfloor\frac {p - 1}{q}\rfloor q \right)^{\lfloor\frac {p - 1}{q}\rfloor q}$ Factoring out a $ q$ from all these terms gives us $ q^{(q + 2q + \cdots + \lfloor\frac {p - 1}{q}\rfloor q )} = q^{\frac {q}{2}\lfloor\frac {(p - 1)}{q}\rfloor \left( \lfloor\frac {p - 1}{q}\rfloor + 1 \right)}$ In all, there are $ \lfloor\frac {p - 1}{q}\rfloor$ multiples of $ q$ in the numerator. Of these $ \lfloor\frac {p - 1}{q^{2}}\rfloor$ are multiples of $ q^{2}$. Factoring out a $ q$ from these terms gives us $ q^{(q^{2} + 2q^{2} + \cdots + \lfloor\frac {p - 1}{q^{2}}\rfloor q^{2})} = q^{\frac {q^{2}}{2}\lfloor\frac {p - 1}{q^{2}}\rfloor \left( \lfloor\frac {p - 1}{q^{2}}\rfloor + 1 \right)}$. Continuing in this fashion, we can compute the exponent of $ q$ in the numerator as follows: $ \frac {q}{2}\lfloor\frac {p - 1}{q}\rfloor \left( \lfloor\frac {p - 1}{q}\rfloor + 1 \right) + \frac {q^{2}}{2}\lfloor\frac {p - 1}{q^{2}}\rfloor \left( \lfloor\frac {p - 1}{q^{2}}\rfloor + 1 \right) + \cdots + \frac {q^{t}}{2}\lfloor\frac {p - 1}{q^{t}}\rfloor \left( \lfloor\frac {p - 1}{q^{t}}\rfloor + 1 \right)$. Comparing the exponent of $ q$ in the numerator and the denominator , it suffices to show $ \frac {(p + 1)}{2}\left(\lfloor\frac {p - 1}{q^{i}}\rfloor \right) \leq \frac {q^{i}}{2}\lfloor\frac {p - 1}{q^{i}}\rfloor \left( \lfloor\frac {p - 1}{q^{i}}\rfloor + 1 \right)$. It is enough to show $ (p + 1) \leq q^{i}\left( \lfloor\frac {p - 1}{q^{i}}\rfloor + 1 \right) \;\text{\;(i.e.)\;}\; \frac {(p + 1)}{q^{i}}\leq \lfloor\frac {p - 1}{q^{i}}\rfloor + 1$. Suppose $ q^{i}$ divides $ p + 1$. Then $ \lfloor\frac {p - 1}{q^{i}}\rfloor = \frac {(p + 1)}{q^{i}} - 1 (i.e.) \frac {(p + 1)}{q^{i}} = \lfloor\frac {p - 1}{q^{i}}\rfloor + 1$. Suppose $ q^{i}$ does not divide $ p + 1$. Then $ \lfloor\frac {(p + 1)}{q^{i}}\rfloor = \lfloor \frac {p - 1}{q^{i}}\rfloor$ (i.e.) $ \frac {(p + 1)}{q^{i}} = \lfloor\frac {p - 1}{q^{i}}\rfloor + f$ where $ f$ is a fraction less than $ 1$. Hence, $ \frac {(p + 1)}{q^{i}}\leq \lfloor\frac {p - 1}{q^{i}}\rfloor + 1$. Therefore, $ Q(p)$ is an integer. In fact, a perfect square.

Nice problem . AMM problem always takes my interest . Solution : Fristly , we will recall an well-known identity : $ \prod_{k = 1}^{n - 1}k! \ = \ \prod_{k = 1}^{n - 1} k^{n - k} \ \ (*) \ \ (n \ge 2)$ Indeed , $ \prod_{k = 1}^{n - 1}k! \ = \ (1 ) . (1.2)....(1.2.....(n - 1))$ we can see that in this product , the numbers $ 1$ apprears $ n - 1$ times , the numbers $ 2$ appears $ n - 2$ times ,...., the numbers $ n - 1$ appears only one time . So , we can deduce that : $ \prod_{k = 1}^{n - 1}k! \ = \prod_{k = 1}^{n - 1}(n - k)! \ = \ 1^{n - 1} . 2^{n - 2} \cdots (n - 1)^1 \ = \ \prod_{k = 1}^{n-1} k^{n - k} \ \ (**)$ , Done Now return to our problem : We can rewrite $ \mathbb{Q} (p)$ as follows : $ \mathbb{Q} (p) = \prod_{k = 1}^{p - 1} k^{2k - p - 1} = \frac {\prod_{k = 1}^{p - 1} k^{k - 1} }{ \prod_{k = 1}^{p - 1} k^{p - k} }$ $ = \frac {\prod_{k = 1}^{p - 1} k^{k - 1} \cdot \prod_{k = 1}^{p - 1} k^{p - k} }{ \left(\prod_{k = 1}^{p - 1} k^{p - k} \right)^2} \ = \ \frac {\prod_{k = 1}^{p - 1} k^{p - 1} }{\prod_{k = 1}^{p - 1}(p - k)! \cdot \prod_{k = 1}^{p - 1}k! }$ ( Using $ (**)$ ) $ = \frac { ((p - 1)!)^{p - 1}}{\prod_{k - 1}^{p - 1} k! (p - k)! } \ = \ \frac {1}{p^{p - 1}} \cdot \frac { (p!)^{p - 1}}{\prod_{k - 1}^{p - 1} k! (p - k)! } = \frac {1}{p^{p - 1}} \cdot \prod_{k = 1}^{p - 1} \frac {p!}{ k! \cdot (p - k)!}$ So , we can have the beautiful identity : $ \mathbb{Q} (p) = \prod_{k = 1}^{p - 1} \frac {\binom{p}{k} }{p}$ We have known that : if $ p$ is a prime , then : $ p | \binom{p}{k} \ \ \forall \ \ k = \overline{ 1 ; p - 1}$ $ \rightarrow \frac {\binom{p}{k} }{p} \in \mathbb{N} \ \ \forall \ \ k = \overline{ 1 ; p - 1} \rightarrow \mathbb{Q} (p) \in \mathbb{N}$ , QED

Let $q$ a prime smaller than $p$, $n=\lfloor{\log_q p}\rfloor$ and $j$ be an integer, $1\leq j\leq n$. We start from this obvious fact: $\left\lfloor{\frac{p}{q^j}}\right\rfloor+1>\frac p{q^j}$. This gives that $q^j\left (\left\lfloor\frac{p}{q^j}\right\rfloor+1\right )$ is strictly greater that $p$, which is to say, greater or equal than $p+1$. Multiplying this inequality by $\left\lfloor{\frac{p}{q^j}}\right\rfloor$ gives: $q^j\left (\left\lfloor{\frac{p}{q^j}}\right\rfloor^2+\left\lfloor\frac{p}{q^j}\right\rfloor \right ) \geq (p+1)\left\lfloor\frac{p}{q^j}\right\rfloor$, or $2q^j\sum\limits_{i=1}^{\left\lfloor\frac{p}{q^j}\right\rfloor} i \geq (p+1)\left\lfloor\frac{p}{q^j}\right\rfloor$ (*) and this is true for every positive integer $j$. We'll define $\mu_q(m)$ for every positive integer $m$ as the greatest integer $r$ such that $q^r$ divides $m$. We shall now compute: $\mu_q\left (\prod\limits_{k=1}^{p-1} k^{2k} \right )=\sum\limits_{k=1}^{p-1} 2k\mu_q(k)=\sum\limits_{j=1}^n \sum\limits_{i=1}^{\left\lfloor\frac{p}{q^j}\right\rfloor} 2ijq^j$. (**) Note that we can extend the inner sum to $\left\lfloor\frac{p}{q^j}\right\rfloor$ (and not only to $\left\lfloor\frac{p-1}{q^j}\right\rfloor$ ) because $q^j$ cannot divide $p$. To finish, we compute $\mu_q\left (\prod\limits_{k=1}^{p-1} k^{p+1} \right )=\sum\limits_{k=1}^{p-1} (p+1)\mu_q(k)=\sum\limits_{j=1}^n \sum\limits_{i=1}^{\left\lfloor\frac{p}{q^j}\right\rfloor} (p+1)j=\sum\limits_{j=1}^n j(p+1)\left\lfloor\frac{p}{q^j}\right\rfloor$. (***) (*), (**) and (***) finish the proof. The sums in $j$ in (*) and (**) may need some further details: the terms of a sum like $\sum\limits_{k=1}^{p-1} f(k)\mu_q(k)$ are 0 when $k$ is not multiple of $q$, $f(k)$ when $k$ is multiple of $q$ but not of $q^2$, $2f(k)$ when $k$ is multiple of $q^2$ but not of $q^3$ and $jf(k)$ when $k$ if multiple of $q^j$ but not of $q^{j+1}$. So, we make $i$ span the numbers that gives less than $p$ when multiplied by $q^j$ and $k=iq^j$.

It is equivalent to prove that, for any prime $q$, \begin{align*} v_q \left(\prod_{k=1}^{p-1}k^{2k}\right)\geq v_q \left(\prod_{k=1}^{p-1}k^{p+1}\right) & \iff 2\sum_{k=1}^{p-1} k v_q(k) \geq (p+1)\sum_{k=1}^{p-1} v_q(k) \\ & \iff 2\sum_{k=1}^{\lfloor p/q \rfloor} k\cdot q+2\sum_{k=1}^{\lfloor p/q^2 \rfloor} k\cdot q^2+2\sum_{k=1}^{\lfloor p/q^3 \rfloor} k\cdot q^3+\dots \geq (p+1)\sum_{k=1}^{\lfloor p/q\rfloor} 1+(p+1)\sum_{k=1}^{\lfloor p/q^2\rfloor} 1+(p+1)\sum_{k=1}^{\lfloor p/q^2\rfloor} 1 +\dots \\ & \iff q \left \lfloor \frac pq \right \rfloor \left(\left \lfloor \frac pq \right \rfloor +1\right)+q^2\left \lfloor \frac p{q^2} \right \rfloor \left(\left \lfloor \frac p{q^2} \right \rfloor +1\right)+\dots \geq (p+1)\left(\left \lfloor \frac pq \right \rfloor+\left \lfloor \frac p{q^2}\right \rfloor+\dots \right).\end{align*}So it suffices to prove for every $n\geq 1$ $$q^n \left (\left \lfloor \frac p{q^n} \right \rfloor+1\right) \geq p+1,$$but this is obvious.