Japan Mathematical Olympiad

$82$, $167$, $1034$

We have $$\frac{(ab-1)(ac-1)}{bc}=\left(a-\frac{1}{b}\right)\left(a-\frac{1}{c}\right).$$ Since we have $0\le a-1\le a-\frac{1}{b}<a$ and $0\le a-1\le a-\frac{1}{c}<a$, we conclude $$(a-1{)}^2\le \frac{(ab-1)(ac-1)}{bc}<a^2,$$ and hence, $(a-1{)}^2\le 2023<a^2$. Since $44^2=1936<2023<2025=45^2$, we have $a-1\le \sqrt{2023}<45$ and $44<\sqrt{2023}<a$. Therefore, we have $44<a<46$, and hence, $a=45$.

By substituting $a=45$ into $\frac{(ab-1)(ac-1)}{bc}=2023$, we get $(45b-1)(45c-1)=2023bc$, which can be rewritten as $2bc-45b-45c+1=0$. From this, we obtain $4bc-90b-90c+2=0$, and hence, we have $(2b-45)(2c-45)=2023$. Since $2023=7\cdot 17^2$ and $-43\le 2b-45\le 2c-45$, the possible pairs for $(2b-45,2c-45)$ are $(1,2023)$, $(7,289)$, and $(17,119)$. Then the corresponding pairs are given by $(b,c)=(23,1034)$, $(26,167)$, $(31,82)$, and all of them satisfy the given conditions.

Therefore, the positive integers $(a,b,c)$ that satisfy the conditions are $(45,23,1034)$, $(45,26,167)$, and $(45,31,82)$. Hence, the answer is $82$, $167$, $1034$.