Japan 2007

Now put $p=n-m$, $q=n$. Finding all positive integer pairs $(m,n)$ is equivalent to finding all integer pairs $(p,q)$. And then condition (1) and (3) is equivalent to the conditions that $1\le q\le 20$, $\frac{1}{4}<\frac{p}{q}<\frac{2}{7}$. And condition (2) is equivalent to the condition that $p$ and $q$ are relatively prime. And $\frac{1}{4}<\frac{p}{q}<\frac{2}{7}$ is equivalent to $\frac{7}{2}p<q<4p$. So, If $p\le 0$, $q$ doesn't exist because of $\frac{p}{q}\le 0$ If $p=1$, then $\frac{7}{2}<q<4$, so $q$ doesn't exist. If $p=2$, then $7<q<8$, so $q$ doesn't exist. If $p=3$, then $\frac{21}{2}<q<12$, so $q=11$. If $p=4$, then $14<q<16$, so $q=15$. If $p=5$, then $\frac{35}{2}<q<20$, so $q=18,19$. * If $p\ge 6$, $q$ doesn't exist because $q>\frac{7}{2}p\ge 21$. Then all integer pairs $(p,q)$ are $(3,11)$, $(4,15)$, $(5,18)$, $(5,19)$. So, all positive integer pairs $(m,n)$ are $(13,18)$, $(8,11)$, $(11,15)$, $(14,19)$.