Asia Pacific Mathematics Olympiad (APMO)

Note that $r=\left(\genfrac{}{}{0ex}{}{p^2}{p}\right)-p$. Hence, it suffices to show that $$\left(p^2-1\right)\left(p^2-2\right)\cdots \left(p^2-(p-1)\right)-(p-1)!\equiv 0\left(\mathrm{mod}{p}^4\right).$$ Now, let $$f(x):=(x-1)(x-2)\cdots (x-(p-1))=x^{p-1}+s_{p-2}{x}^{p-2}+\cdots +s_{1}x+s_0.$$ Then the congruence equation (1) is the same as $f\left(p^2\right)-s_0\equiv 0(\mathrm{mod}{p}^4)$. Therefore, it suffices to show that $s_1{p}^2\equiv 0(\mathrm{mod}{p}^4)$ or $s_1\equiv 0(\mathrm{mod}{p}^2)$.

Since $a^{p-1}\equiv 1(\mathrm{mod}p)$ for all $1\le a\le p-1$, we can factor $$x^{p-1}-1\equiv (x-1)(x-2)\cdots (x-(p-1))(\mathrm{mod}p)$$ Comparing the coefficients of the left hand side with those of the right hand side, we obtain $p\mid s_i$ for all $1\le i\le p-2$ and $s_0\equiv -1(\mathrm{mod}p)$. On the other hand, plugging $p$ for $x$ in the expansion, we get $$f(p)=(p-1)!=p^{p-1}+s_{p-2}{p}^{p-2}+\cdots +s_{1}p+s_0$$ which implies $$p^{p-1}+s_{p-2}{p}^{p-2}+\cdots +s_2{p}^2=-s_{1}p$$ Since $p\ge 5$, $p\mid s_2$ and hence $s_1\equiv 0(\mathrm{mod}{p}^2)$ as desired.