Solution: We can see that, for all integers $p,q$, we will have $p^2+q^2+2020>3$. If $p,q>2$, then $p^2+q^2+2020\equiv 0$ mod $2$, and is thus composite. Also, if $p=q=2$, then $p^2+q^2+2020\equiv 0$ mod $2$ and is composite. WLOG $p=2$ and $q>2$. Then ,if $q\equiv 1,2$ mod $3$, we will have $q^2\equiv p^2\equiv1$ mod $3$, and $p^2+q^2+2020\equiv 1+1+1\equiv 0$ mod $3$, so it's composite. Now consider $p=2$ and $q=3$. Then $p^2+q^2+2020=4+9+2020=2033=19\cdot107$, so for all primes $p,q$, $p^2+q^2+2020$ will be composite.

We know that any square integer is of the form 3k or 3k+1 Case-I We have p,q as prime, so let both p & q be even prime i.e. 2. Now, $ p^2 + q^2 +2020=2028$. Case-II Let one of them be 2 and $p>3$. Then $p^2 + q^2 +2020= p^2 + 2024$. Now observe that $2024 \equiv -1(mod 3)$ and $p^2 \equiv 1(mod3)$. Hence the equation becomes divisible by 3. Case-IIIWe can easily check the case when $q=3$ and $p=2$, so we consider the case when $p,q>3$. The equation becomes $p^2+ q^2+2020 \equiv 1(mod3) + 1(mod3) + 1(mod3) \equiv 0(mod3)$. Hence proved $\blacksquare$

Try $mod$ $3$

Note that for any prime $p>3, p^2 \equiv 1 \pmod{3}$ If $p,q$ are both even or odd, it is clear that $p^2+q^2+2020$ is even, and therefore composite We have 2 other cases: Case 1: $p=2, q=3 \Rightarrow p^2+q^2+2020=2033$, which is composite Case 2: $p=2, q>3 \Rightarrow p^2+q^2+2020=q^2+2024 \equiv 1 \pmod{3} + 2 \pmod{3} \equiv 0 \pmod{3} \Rightarrow 3|p^2+q^2+2020$ Therefore, $p^2+q^2+2020$ is composite $\blacksquare$