Assume ${\textstyle \frac{(2n)! + 1}{n! + 1} \in \mathbb{N}}$. Then $\frac{(2n)!}{n!} - \frac{(2n)! + 1}{n! + 1} = \frac{(2n)! - n!}{n!(n! + 1)} = \frac{2n(2n - 1) \cdots (n + 1) - 1}{n! + 1} = M_n \in \mathbb{N}$. Hence ${2n \choose n} - M_n = F(n) \in \mathbb{Z}$, i.e. $(1) \;\; F(n) = \frac{(2n)!}{(n!)^2} - \frac{(2n)! - n!}{n!(n! + 1)} = \frac{(2n)!(n! + 1) - ((2n)! - n!)n!}{(n!)^2(n! + 1)} = \frac{(2n)! + (n!)^2}{(n!)^2(n! + 1)} \in \mathbb{N}$. Hence by (1) $(2) \;\; 0 < F(n) < \frac{2 \cdot (2n)!}{(n!)^3} = G(n)$. Consequently $\frac{G(n+1)}{G(n)} = \frac{2(2n + 1)}{(n + 1)^2}$, implying $G(n+1) < G(n)$ when $n \geq 3$. Hence, if $n \geq 8$, then ${\textstyle 0 < F(n) < G(n) \leq G(8) = \frac{143}{224} < 1}$. Therefore ${\textstyle \frac{(2n)! + 1}{n! + 1} \in \mathbb{N}}$ only if $n \leq 7$. Checking these seven values of $n$, we end up with this Conclusion: ${\textstyle \frac{(2n)! + 1}{n! + 1} \in \mathbb{N}}$ iff $n=3$.

Solar Plexsus wrote: Assume ${\textstyle \frac{(2n)! + 1}{n! + 1} \in \mathbb{N}}$. Then $\frac{(2n)!}{n!} - \frac{(2n)! + 1}{n! + 1} = \frac{(2n)! - n!}{n!(n + 1)!} = M_n \in \mathbb{N}$, implying $n! \cdot M_n = \frac{(2n)! - n!}{(n + 1)!} = \frac{(2n)!}{(n + 1)!} - \frac{1}{n + 1} \in \mathbb{N}$. Hence ${\textstyle \frac{1}{n + 1} \in \mathbb{N}}$, which obviously is false since $n$ is a natural number. Thus we have proven by contradiction that ${\textstyle \frac{(2n)! + 1}{n! + 1} \not \in \mathbb{N}}$ for all natural numbers $n$. i think something is wrong, as $\frac{(2n)!}{n!} - \frac{(2n)! + 1}{n! + 1} = \frac{(2n)! - n!}{n!(n! + 1)} \ne \frac{(2n)! - n!}{n!(n + 1)!}$

mathisreal wrote: Find all positive integers $n$ such that the number $\frac{(2n)!+1}{n!+1}$ is positive integer. Is this an actual contest problem? It seems much difficult.

Asume $\frac{(2n)!+1}{n!+1}$ is positive inter then $n!+1$ divide $(2n)!+1$ => $n!+1$ divide $(n+1)(n+2).......(2n)-1$ let $(n+1)(n+2).......(2n)-1$=$k.(n!+1)$ mod $n!$ => $k+1$ is divided by $n!$ => $k.(n!+1) \ge (n!)^2-1$ we can check $(n!)^2> (n+1)(n+2)......2n$ for $n \ge 7 $ and easy to prove by induction check results for n= 1 2 3 4 5 6

Assume $X_{n}=\frac{(2n)!+1}{n!+1} \in \mathbb{N}$. Note that for all positive integers $n$ we have that $gcd(n!,n!+1)=1$. \[n!+1|(2n)!+1\Rightarrow n!+1|((2n)!+1)-(n!+1) \Rightarrow n!+1|(2n)!-n! \]\[\Rightarrow n!+1|\frac{(2n)!}{n!}-1 \Rightarrow n!+1|\frac{(2n)!}{n!}-1+n!+1\Rightarrow n!+1|\frac{(2n)!}{n!}+n!\Rightarrow n!+1|\frac{2n!}{(n!)^2}+1\]\[\Rightarrow n!+1|{2n\choose n}+1\] Hence ${2n \choose n} \geq n!$. But we have that ${2n\choose n} = \frac{(2n-1)!!\cdot 2^n\cdot n!}{(n!)^2}=2^n\frac{(2n-1)!!}{n!}=2^n \prod_{i=1}^{i=n} \frac{2i-1}{i} \leq 2^n \prod_{i=1}^{i=n} 2 =2^{2n}$ Thus $n!\leq {2n \choose n}\leq 2^{2n}=4^{n}$ It is clear that the left hand side grows faster than the right hand side for $n\geq 4$, thus when $n!> 4^{n}$ for $n\geq 4$, we have that $N!> 4^{N}$ $\forall$ $N\geq n$. It is easy to see that $9!>4^9$, and thus we only need to consider the cases $n=1,2,3,4,5,6,7,8$. It turns out that only $n=3$ yields an integer, and thus our answer must be $n=3$. $\blacksquare$

kwanglee123456 wrote: Asume $\frac{(2n)!+1}{n!+1}$ is positive inter then $n!+1$ divide $(2n)!+1$ => $n!+1$ divide $(n+1)(n+2).......(2n)-1$ let $(n+1)(n+2).......(2n)-1$=$k.(n!+1)$ mod $n!$ => $k+1$ is divided by $n!$ => $k.(n!+1) \ge (n!)^2-1$ we can check $(n!)^2> (n+1)(n+2)......2n$ for $n \ge 7 $ and easy to prove by induction check results for n= 1 2 3 4 5 6 You are right. I missed such a typical trick

I have corrected the error the signature Aniv pointed out.