The answer is $(p,q)=(-6,16) (-6,-16) (-10,0)$ Key observation is to prove that $144=(ab-cd)^2+(ac-bd)^2+(ad-bc)^2$, which $a,b,c,d$ are four roots of the equation

This was like an calculus problem; my solution involves Rolle's theorem and some caseworks.

Let $a,b,c,d$ be the four roots of the equation that $a\equiv b\pmod{2}$. $a+b+c+d=0$ gives us $c\equiv d\pmod{2}$. Also WLOG assume $a^2\geqslant b^2,c^2\geqslant d^2,$ and $ab\geqslant cd$. We have $ab+cd-(a+b)^2=2p$ and $abcd=p^2-36$. This gives $(ab+cd)^2-2(ab+cd)(a+b)^2+(a+b)^4-4abcd=144$. Note that $$(ab+cd)^2-2(ab+cd)(a+b)^2+(a+b)^4-4abcd = (ab-cd)^2+(a+b)^2((a+b)^2-2(ab+cd))=(ab-cd)^2+(a+b)^2(a^2+b^2-2cd).$$From $(a+b)^2=(c+d)^2$, we get $$a^2+b^2-2cd=c^2+d^2-2ab\implies a^2+b^2-2cd=\frac{(a^2+b^2-2cd)+(c^2+d^2-2ab)}{2}=\frac{(a-b)^2+(c-d)^2}{2}.$$Hence, we get $$144=(ab-cd)^2+(a+b)^2\frac{(a-b)^2+(c-d)^2}{2}\implies 288=2(ab-cd)^2+(a^2-b^2)^2+(c^2-d^2)^2.$$We have $4\mid a^2-b^2,c^2-d^2\implies 16\mid 2(ab-cd)^2\implies 4\mid ab-cd$. So, $$18=2\left( \frac{ab-cd}{4}\right)^2 +\left( \frac{a^2-b^2}{4}\right)^2 +\left( \frac{c^2-d^2}{4}\right)^2.$$Clearly, the only non-negative integer solutions $(x,y,z)$ to $18=2x^2+y^2+z^2$ are $$(0,3,3),(1,4,0),(1,0,4),(2,3,1),(2,1,3),(3,0,0).$$From here the rest is easy.

My solution Let the four roots be $a,b,c,d$. $a+b+c+d=0$, $ab+bc+cd+da+ac+bd=2p$, $abcd=p^2-36$. Just substitute $d=-a-b-c$, and we get $\left(-a^2-b^2-c^2-ab-bc-ca\right)=2p$, $-abc(a+b+c)=p^2-36$. Now let $x=b+c$, $y=c+a$, $z=a+b$. Then we get $x^2+y^2+z^2=-4p$, $2\left(y^2z^2+z^2x^2+x^2y^2\right)-\left(x^4+y^4+z^4\right)=36-p^2$ Add the square of the first equation to the second equation , then $y^2z^2+z^2x^2+x^2y^2=144$. After analyzing by $\mod 2$, we get that $x,y,z$ are all even. $\therefore$ $\left(\frac{y}{2}\right)^2\left(\frac{z}{2}\right)^2+\left(\frac{z}{2}\right)^2\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2\left(\frac{y}{2}\right)^2=9$. Now it's just a matter of casework.

By the way, I would like to see solution from outline in #3, can you provide more details? lminsl wrote: This was like an calculus problem; my solution involves Rolle's theorem and some caseworks.