It is known that $$ x^{2} - 2px -5q = (x-m)(x-n)$$ and $$ x^{2} - 2mx -5n = (x-p)(x-q)$$
with $p \ne q \ne m \ne n$. What is $p+q+m+n$?
Attempt:
We get that $$ m+n = 2p, \:\:\: mn = -5q $$ $$ p +q = 2m, \:\: pq = -5n $$
We can get that $$ (m+n)(p+q) = 100 $$ from $ m + n = 2 (-5n/q) $ and $ p +q = 2(-5q/n) $.
So $$ ((p + q) + (m + n))^{2} = (m+n)^{2} + (p+q)^{2} + 200$$
We also can get $n+q = m + p$. So
$$ ((n + q) + (m + p))^{2} = 4 (m+p)^{2} = 4 (n+q)^{2}$$
How to continue?
We need to solve the following system: $$m+n=2p,$$ $$p+q=2m,$$ $$mn=-5q$$ and $$pq=-5m.$$ Since $n=2p-m$ and $q=2m-p,$ we obtain: $$m(2p-m)=-5(2m-p)$$ and $$p(2m-p)=-5(2p-m),$$ which gives $$m(2p-m)-p(2m-p)=-5(2m-p)+5(2p-m)$$ or $$p^2-m^2=15p-15m$$ or $$(p-m)(p+m-15)=0.$$ Can you end it now?