It is clear that $p=2$ is not a solution. So, we assume $p$ is not $2$. In that case, LHS is divisible by $4$. This implies $q^2 +r^2+s^2$ is divisible by $2$ which is impossible when all $3$ are odd. WLOG, assume $q=2$. Now, consider $mod 3$. We see that if none of $r,s$ is $3$, we have RHS divisible by $3$. So, $p=3$ gives $r=5$ and $s=7$. Otherwise one of $r,s$ is $3$. This should be easy now using mod 5 argument as test20 below does.

I continue and complete the argument of @meet18: In the only remaining case, w.l.o.g. we have $q=2$ and $r=3$ and $p^2+2019=26(2^2+3^2+s^2)$. Now working modulo $5\,$, the equation becomes $s^2=p^2+1$. Since modulo $5$ the only squares are $0,\pm1$, this implies that one of $s$ and $p$ is divisible by $5$, and as a prime hence equal to $5$. Neither of these two options yields a solution of the original equation.

i guess,,,,,

jbaca wrote: Find all positive prime numbers $p,q,r,s$ so that $p^2+2019=26(q^2+r^2+s^2)$.