Team Selection Test

Using the condition given, we have $(m+n{)}^2+(2018-m+n{)}^2=2018^2$. Let $m+n=u$, $2018-m+n=v$. It can be seen that $u,v>0$. Since $u^2+v^2=2018^2\equiv 0(\mathrm{mod}4)$, we get $u\equiv v\equiv 0(\mathrm{mod}2)$. Let $u=2u_1$, $v=2v_1$. Then we have $u_1^2+v_1^2=1009^2$. Since $(u_1,v_1,1009)$ is a Pythagorean Triple, we get $1009=d(r^2+s^2)$, $(u_1,v_1)=(d(2rs),d(r^2-s^2))$ or $(u_1,v_1)=(d(r^2-s^2),d(2rs))$ for some positive integers $d,r,s$. Since $1009$ is a prime number, we obtain $d=1$ and $r^2+s^2=1009$. The last equation has solutions $(r,s)=(28,15),(15,28)$. Hence all solutions are $(u_1,v_1)=(840,559),(559,840)$ which correspond $(m,n) = (728,390), (1290,390).