It's easy to know $n=1,2,3,4$,$\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor$ is even. Now assume $n\geq5$: (1)$n$ is odd prime,By Wilson:$(n-1)!+1\equiv0\pmod{n}$ And $n+1$ is composite,suppose $n+1=ab,2\leq a\leq b\leq n-1$ (i)$a\neq b$,$a,b\in\{1,2,\ldots,n-1\},\therefore n+1=ab|(n-1)!$ (ii)$a=b$,$n+1=a^2(a\geq3),a,2a\in\{1,2,\ldots,n-1\},n+1|(n-1)!$ So $n(n+1)|(n-1)!+n+1$,and $n+1<n(n+1)$ $\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n+1}{n(n+1)}-1=\frac{(n-1)!-(n-1)(n+1)}{n(n+1)}$ Since 2 cannot divide n,suppose $2^k||n+1,n+1\geq2^k$ We prove $2^{k+1}|(n-1)!-(n-1)(n+1)$ Since $2|n-1,2^{k+1}|(n-1)(n+1)$. And $2^t||(n-1)!,t=\sum_{i=1}^{\infty}\left\lfloor\frac{n-1}{2^i}\right\rfloor\geq\left\lfloor\frac{2^k-2}{2^i}\right\rfloor=2^{k-1}-1+2^{k-2}-1+\cdots+2-1=2^k-k-1\geq k+1$ Easy to know it's correct when $k\geq 3$ And when $k=1,2$,$n+1\geq6,n-1\geq4,8|(n-1)!,t\geq3,t\geq k+1$ So $\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor$ is even.

(2)$n+1$ is odd prime,$n!+1\equiv0\pmod{n+1},(n-1)!\equiv1\pmod{n+1}$ Similar with (1)$n|(n-1)!$ $n(n+1)|(n-1)!+n,\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n}{n(n+1)}-1=\frac{(n-1)!-n^2}{n(n+1)}$ Similar with (1),$\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor$ is even. (3)$n,n+1$ is composite,$n,n+1$ is relatively prime, $n(n+1)|(n-1)!,\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!}{n(n+1)}$ Similar with (1),$\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor$ is even.

shobber wrote: For a real number $x$, let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that \[ \left\lfloor{(n-1)!\over n(n+1)}\right\rfloor \] is even for every positive integer $n$. Check the result for $n \le 8$. For $n \ge 8$ we consider three cases. Assume first $n$ and $n+1$ are not prime. We find that $n \mid (n-1)!$ and $n+1 \mid (n+1)!$, and hence $\frac{(n-1)!}{n(n+1)}$ is an integer; simple bounding on the $2$-adic valuation shows that it is even. Assume next that $n = p$ is prime, and $p \ge 11$. Evidently $\frac{(p-1)!}{p+1} \equiv -1 \pmod{p}$ (by Wilson) and accordingly the quantity is \[ -1 + \frac{\frac{(p-1)!}{p+1}+1}{p} \] which is easily seen to be even. Finally assume $n = p-1$, where $p \ge 11$ is prime. Now $\frac{(p-2)!}{p-1} \equiv -1 \pmod{p}$ and accordingly the quantity is now \[ -1 + \frac{\frac{(p-2)!}{p-1}+1}{p} \] which is also easily seen to be even.

For the case where $n$ and $n+1$ are composite, after showing that $\left\lfloor{\frac{(n-1)!}{n(n+1)}}\right\rfloor={\frac{(n-1)!}{n(n+1)}}$, to show that this is even, is it enough to say that $\displaystyle\sum_{i=1}^\infty{\left\lfloor\frac{(n-1)!}{2^i}\right\rfloor}$ is clearly greater than $\text{max}\left\{\displaystyle\sum_{i=1}^\infty{\left\lfloor\frac{n}{2^i}\right\rfloor}, \displaystyle\sum_{i=1}^\infty{\left\lfloor\frac{n+1}{2^i}\right\rfloor}\right\}$? Like can I just state it as obvious?

Well $v_2((n-1)!) = n-1-s_2(n-1) \ge n-\log_2(n-1)$ by the bound $s_2(n-1)\le \log_2(n-1)+1$. However, $v_2(n(n+1)) < \log_2(n)+\log_2(n+1)$ so it remains to show that $$n-\log_2(n-1) \ge \log_2(n)+\log_2(n+1)+1\implies n\ge \log_2(n)+\log_2(n+1)+\log_2(n-1)+1\implies 2^{n-1}\ge n(n+1)(n-1)$$which is clearly true for $n\ge 12$ so you just need to check $n=1\to 11$ and since $n, n+1$ are not prime we just need to check $n=8, 9$.

v_Enhance wrote: shobber wrote: For a real number $x$, let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that \[ \left\lfloor{(n-1)!\over n(n+1)}\right\rfloor \]is even for every positive integer $n$. Check the result for $n \le 8$. For $n \ge 8$ we consider three cases. Assume first $n$ and $n+1$ are not prime. We find that $n \mid (n-1)!$ and $n+1 \mid (n+1)!$, and hence $\frac{(n-1)!}{n(n+1)}$ is an integer; simple bounding on the $2$-adic valuation shows that it is even. Assume next that $n = p$ is prime, and $p \ge 11$. Evidently $\frac{(p-1)!}{p+1} \equiv -1 \pmod{p}$ (by Wilson) and accordingly the quantity is \[ -1 + \frac{\frac{(p-1)!}{p+1}+1}{p} \]which is easily seen to be even. Finally assume $n = p-1$, where $p \ge 11$ is prime. Now $\frac{(p-2)!}{p-1} \equiv -1 \pmod{p}$ and accordingly the quantity is now \[ -1 + \frac{\frac{(p-2)!}{p-1}+1}{p} \]which is also easily seen to be even. Can someone explain the following : We find that $n \mid (n-1)!$ and $n+1 \mid (n+1)!$, and hence $\frac{(n-1)!}{n(n+1)}$ is an integer; simple bounding on the $2$-adic valuation shows that it is even.

My solution is quite similar to v-enhance, and p-adic evaluation is looking at $v_p$'s One can check the result for $n < 8$ For n $\geq$ 8, we consider three cases: n, n+1 are composite, n is a prime, and n+1 is a prime. I'll just prove cases 1 and 2 since 3 is quite similar to 2. Case 1: n, n+1 are both composite. Note that n $\mid$ (n-1)! and n+1 $\mid$ (n-1)!. So it suffices to prove that $ \left\lfloor{(n-1)!\over n(n+1)}\right\rfloor $ is even. To do this, we note that that one of n, n+1 must be even and the other odd since gcd(n,n+1) = 1. Hence, the parity of the expression in this case follows from $ log_2(n) < log_2(n+1)$ < $ \left\lfloor{(n-1)!\over 2}\right\rfloor $+ $ \left\lfloor{(n-1)!\over 4}\right\rfloor $ < $v_2(n-1)!$ Case 2: n=p is prime n+1=p+1 is composite. By wilson's theorem, note that $\frac{(p-1)! + p + 1}{p(p+1)}$ is an integer. So $ \left\lfloor{(p-1)!\over p(p+1)}\right\rfloor $ = $\frac{(p-1)! + p + 1}{p(p+1)}$ - 1. So it's sufficient to show that $\frac{(p-1)! + p + 1}{p(p+1)}$ is odd. Divide both the numerator and denominator by p+1 to get: $\frac{\frac{(p-1)!}{p+1} + 1}{p}$ Since $\frac{(p-1)!}{p+1}$ is even (follows from case 1), $\frac{(p-1)!}{p+1}+1$ is odd. Hence, $\frac{\frac{(p-1)!}{p+1} + 1}{p}$ is odd. So case 2 is even. So we proved that case 2 and case 1 is even, done.

https://imgur.com/a/sfexK Edit: how do I insert an image? I tried but it isn't working

This is clearly true by $n\leq 9$ by manual verification. Assume $n\geq10$ from now on. For $n\geq10$, as long as neither $n$ nor $n+1$ are prime, then clearly the expression inside the floor is already an integer. Furthermore, $(n-1)!$ contains far more factors of 2 than $n(n+1)$, so it will be even. Case 1: $n$ is prime. Let $n=p$. Then, it becomes $$\lfloor \frac{(p-1)!}{p(p+1)}\rfloor.$$Clearly, $(p-1)!/(p+1)$ is an integer as $p+1$ is not prime. Now, note that $$(p-1)!\equiv -1\pmod p$$by Wilson, so $$\frac{(p-1)!}{p+1}\equiv\frac{-1}{1}\equiv-1\pmod p.$$Thus, $$\lfloor \frac{(p-1)!}{p(p+1)}\rfloor=\frac{\frac{(p-1)!}{p+1}-(p-1)}{p}$$$$\frac{(p-1)!-(p-1)(p+1)}{p(p+1)}.$$Clearly, $v_2((p-1)!)>>>v_2((p-1)(p+1))$ since $p\geq11$, so the $v_2$ of the numerator is just $v_2((p-1)(p+1))$. This is clearly greater than $v_2(p(p+1))$, as $p-1$ is even while $p$ is odd, so this is even and this case is resolved. Case 2: $n+1$ is prime. Let $n=p-1$. Then, it becomes $$\lfloor \frac{(p-2)!}{p(p-1)}\rfloor.$$Note that $$(p-2)!\equiv 1\pmod{p},$$so $$\frac{(p-2)!}{p-1}\equiv -1\pmod{p}.$$Thus, $$\lfloor \frac{(p-2)!}{p(p-1)}\rfloor=\frac{\frac{(p-2)!}{p-1}-(p-1)}{p}$$$$=\frac{(p-2)!-(p-1)^2}{p}.$$The numerator is clearly even while the denominator is odd, so this case is also resolved and we are done.

Just writing up my solution from mock JMO a while ago Solved this on the mock JMO in half an hour Suppose n>7 (manually check smaller ones), and if n,n+1 are composite, then $\frac{(n-1)!}{n(n+1)}$ is easily checked to be an even integer. It suffices to only consider when n or n+1 is prime. If n=p a prime, then $\frac{(p-1)!}{p+1}\stackrel{\text{Wilson}}{\equiv}-1\pmod p\implies\frac{(p-1)!}{p+1}=kp-1$ where k is obviously odd, hence divided by p we get $\lfloor\frac{kp-1}{p}\rfloor=k-1$ which is even. If n+1=p a prime, then $\frac{(p-1)!}{(p-1)^2}\equiv-1\pmod p$, and the same reasoning we get it's even. @below yeah its otis version

huashiliao2020 wrote: Just writing up my solution from mock JMO a while ago Solved this on the mock JMO in half an hour Suppose n>7 (manually check smaller ones), and if n,n+1 are composite, then $\frac{(n-1)!}{n(n+1)}$ is easily checked to be an even integer. It suffices to only consider when n or n+1 is prime. If n=p a prime, then $\frac{(p-1)!}{p+1}\stackrel{\text{Wilson}}{\equiv}-1\pmod p\implies\frac{(p-1)!}{p+1}=kp-1$ where k is obviously odd, hence divided by p we get $\lfloor\frac{kp-1}{p}\rfloor=k-1$ which is even. If n+1=p a prime, then $\frac{(p-1)!}{(p-1)^2}\equiv-1\pmod p$, and the same reasoning we get it's even. huashiliao2020 wrote: Just writing up my solution from mock JMO a while ago Solved this on the mock JMO in half an hour Suppose n>7 (manually check smaller ones), and if n,n+1 are composite, then $\frac{(n-1)!}{n(n+1)}$ is easily checked to be an even integer. It suffices to only consider when n or n+1 is prime. If n=p a prime, then $\frac{(p-1)!}{p+1}\stackrel{\text{Wilson}}{\equiv}-1\pmod p\implies\frac{(p-1)!}{p+1}=kp-1$ where k is obviously odd, hence divided by p we get $\lfloor\frac{kp-1}{p}\rfloor=k-1$ which is even. If n+1=p a prime, then $\frac{(p-1)!}{(p-1)^2}\equiv-1\pmod p$, and the same reasoning we get it's even. ... mock jmo sus

bro why is this problem so bashy First, for $n, n + 1$ both composite, we have $n + 1 = ab, n = cd$ with $n - 1 \ge a, b, c, d > 1$, and $a, b, c, d$ all distinct (because $\gcd(n, n + 1) = 1$), so $n(n + 1) \mid (n - 1)!$, so our floor expression is just $\frac{(n - 1)!}{n(n + 1)}$, so it suffices to show that $\nu_2((n - 1)!) > \nu_2(n) + \nu_2(n + 1),$ or that $n - 1 - s_2(n - 1) > \nu_2(n) + \nu_2(n + 1),$ or that $n - 1 > s_2(n - 1) + \nu_2(n) + \nu_2(n + 1)$. But, for $n \ge 16$, we have $n - 1 > 3 \log_2(n + 1) > s_2(n - 1) + \nu_2(n) + \nu_2(n + 1)$, so it suffices to check $n < 16$. Now, the only $n$ we must check are $n = 15, 9, 8,$ since those are only $n$ for which $n, n + 1$ are both composite. For $n = 15,$ we have $\nu_2(n) + \nu_2(n + 1) = 2$, and $n - 1 - s_2(n - 1) = 11,$ so it holds. For $n = 9$, we have $\nu_2(n) + \nu_2(n + 1) = 1,$ and $n - 1 - s_2(n - 1) = 7,$ so it once again holds. For $n = 8,$ we have $\nu_2(n) + \nu_2(n + 1) = 3,$ and $n - 1 - s_2(n - 1) = 4,$ so it once again holds. So, this case is complete. Now, let us assume that either of $n$ or $n + 1$ is prime. Check $n \le 7$ manually, and henceforth assume $n \ge 8$. The first case: $n$ is prime. Thus, we have $\left \lfloor \frac{(p - 1)!}{p(p + 1)} \right \rfloor.$ Now, $(p - 1)! \equiv 0 \pmod{p + 1},$ and $(p - 1)! \equiv -1 \pmod{p}$ by Wilson's, so by CRT, we have unique residue modulo $p(p + 1)$ which can be found to be $p^2 - 1$, so our floor is $\left \lfloor \frac{(p - 1)! - (p^2 - 1)}{p(p + 1)} \right \rfloor$. So, it suffices to show that $\nu_2((p - 1)! - (p^2 - 1)) > \nu_2(p) + \nu_2(p + 1) = \nu_2(p + 1)$ (since $p \ge 8$ is odd prime). But, since $(p + 1) \mid (p - 1)!$, so letting $p + 1 = ab$, where $a, b \le p - 1$, there is at least one another factor of $2$ in $(p - 1)!$ (except $a$ and $b$), since there are at least $4$ even numbers $\le p - 1$, and this accounts for only at most $2$. So, $\nu_2((p - 1)!) > \nu_2(p + 1)$. Also, $\nu_2(p^2 - 1) = \nu_2(p - 1) + \nu_2(p + 1) \ge 1 + \nu_2(p + 1) > \nu_2(p + 1)$, so $\nu_2((p - 1)! - (p^2 - 1)) > \nu_2(p(p + 1))$ for this case as desired. For the case $n = p - 1$, we have $\left \lfloor \frac{(p - 2)!}{p(p - 1)} \right \rfloor$, so similarly to last case, we have $(p - 2)! \equiv 0 \pmod{p - 1}, (p - 2)! \equiv 1 \pmod{p}$, so $(p - 2)! \equiv (p - 1)^2 \pmod{p(p - 1)}$, so our expression is $\frac{(p - 2)! - (p - 1)^2}{p(p - 1)},$ so it suffices to prove that $\nu_2((p - 2)! - (p - 1)^2) > \nu_2(p) + \nu_2(p - 1) = \nu_2(p - 1)$, which can be dealt with just like last time since $\nu_2((p - 1)^2) = 2\nu_2(p - 1) > \nu_2(p - 1)$, and similarly to last time, $\nu_2((p - 2)!) > \nu_2(p - 1)$ since we can cancel out at most $2$ factors. So, we checked all cases, we're done.