Given $p(x)=x^2+ax+b$, where $a,b\in \mathbb{Z}$. If $p(2020 + p(2020))=1$, find $a,b$ 
Given $p(x)=x^2+ax+b$, where $a,b\in \mathbb{Z}$. If $p(2020 + p(2020))=1$,
find $a,b.$

My attempt:
Given $p(x) = x^2+ax+b$, then $p(x+p(x))=p(x^2+(a+1)x+b)=(x^2+(a+1)x+b)^2+a(x^2+(a+1)x+b)+b=1$
After a few pages of algebra, I get...
$(a x + b + x^2) (a x + a + b + x^2 + 2 x + 1) =1$
Substituting $x=2020.$
$(2020a + b + 2020^2) (2020a + a + b + 2020^2 + 2(2020) + 1) =1$
Equating the factors of 1 yields 2 systems of equations.
$$\begin{cases} (2020a + b + 2020^2)=1 \\  (2020a + a + b + 2020^2 + 2(2020) + 1) =1\\ \end{cases}\implies \boxed{a=-4041, b=4082421}$$
$$\begin{cases} (2020a + b + 2020^2)=-1 \\  (2020a + a + b + 2020^2 + 2(2020) + 1) =-1\\ \end{cases}\implies \boxed{a=-4041, b=4082419}$$
This was a competition question. I felt my solution isn't elegant since it required a lot of algebra. Any thoughts on a better solution?

weird fact about parabolas:
$$p(x)=x^2+ax+b\iff \boxed{p(n)p(n+1)}=p\Big(p(n)+n\Big)$$
$$\begin{cases}2020a+b+2020^2&=\pm1\\2021a+b+2021^2&=\pm1\end{cases}$$
$0=2021^2-2020^2+a \iff a=-4041$
$b=\pm 1-2021(-4041)-2021^2=4082420 \pm 1$
