My solution: It is obvious that $(p!,t)$=1，and it's easy to check when$p \leqslant 7$ When $p \geqslant 8$

@above meant $p \leq2q$ in case (2).

Anonimous wrote: If $p \geqslant 2q$,then$(p-q)!|q!$ then $t=\gcd(\frac{p!}{q!}-1,q!-1)=\gcd(\frac{p!}{q!}-q!,q!-1)=\gcd(\frac{p!}{q!(p-q)!}-\frac{q!}{(p-q)!},q!-1)=\gcd(\binom{p}{q}-\frac{q!}{(p-q)!},q!-1)$ so $t \leqslant \binom{p}{q}<2^p\leqslant p^{\frac{p}{3}}$ But $\binom{p}{q}-\frac{q!}{(p-q)!}$ is not necessarily positive.

Rickyminer wrote: But $\binom{p}{q}-\frac{q!}{(p-q)!}$ is not necessarily positive. Thanks! And I have fixed it!

Not to be picky, but it does not quite suffice to just write absolute values. After all, if $\binom{p}{q}=\frac{q!}{(p-q)!}$, then your argument would be just false. Of course, this cannot happen since $p$ is prime, so $p$ divides the LHS but not the RHS. But whenever one uses the argument that $d \mid n$ implies $d \le \vert n\vert$ one should be careful to check that $n \ne 0$. On a different note, does anyone know if the exponent $\frac{1}{3}$ is sharp? That is, can we perhaps prove the bound $p^{cp}$ for some $c<\frac{1}{3}$, at least for all sufficiently large $p$?

Suppose on the contrary that $t>p^{\frac{p}{3}}$, then since $q!,\frac{p!}{q!}>t$, we have $$\frac{p}{3}\leq q\leq \frac{2p}{3}$$ Notice that $(q+1)...p\equiv 1\pmod t$, hence $$t|\frac{p!}{q!}-q!$$let $d=(\frac{p!}{q!},q!)$, define $a=\frac{p!}{dq!}$, $b=\frac{q!}{d}$. Notice that we have $(d,t)=1$, hence $$t|a-b$$and so $$t\leq |a-b|< \max\{a,b\}\hspace{20pt}(1)$$We divide into two cases, Case I: $q\leq \frac{p}{2}$ The key observation is that $$q!|\frac{(2q)!}{q!}$$since $\binom{2q}{q}$ is an integer, hence $q!|\frac{p!}{q!}$, this implies $d=q!$, and so $b=1$. Notice that $d=q!>t$, hence $$t^2<dt< da=\frac{p!}{q!}<p^{\frac{2p}{3}}$$as desired. Case II; $q\geq\frac{p}{2}$ Similarly we have $$(p-q)!|\frac{p!}{q!}$$so $d>(p-q)!$, we have $$a<\binom{p}{q}$$and $$b<\prod_{\frac{p}{3}<x<\frac{2p}{3}}x<p^{\frac{p}{3}}$$For sufficiently large $p$ we have $2^p<p^{\frac{p}{3}}$, so $t<b<p^{\frac{p}{3}}$ as desired.

Is $p$ prime necessary? Can $p$ be replaced by other integers ?

nEpg wrote: Is $p$ prime necessary? Can $p$ be replaced by other integers ? The proposer of the problem has done some researches on what you have mentioned. If you understand Chinese, you can download it at the following website: http://www.nsmath.cn/jszl the paper is on page 5, titled "一道CGMO试题的加强与命题背景", written by Mr. LaiLi Unfortunately, because the file is too large, I cannot upload it as an attachment

Rushery_10 wrote: nEpg wrote: Is $p$ prime necessary? Can $p$ be replaced by other integers ? The proposer of the problem has done some researches on what you have mentioned. If you understand Chinese, you can download it at the following website: http://www.nsmath.cn/jszl the paper is on page 5, titled "一道CGMO试题的加强与命题背景", written by Mr. LaiLi Unfortunately, because the file is too large, I cannot upload it as an attachment Thank you ,I've found the article now.