$4a^{2}+36a+2020=(2a)^{2}+2.2a.9+9^{2}+1939=(2a+9)^{2}+1939$.
Giả sử $4a^{2}+36a+2020=n^{2}=(2a+9)^{2}+1939 ⇔ (2a+9)^{2}-n^{2}=1939 ⇔ (2a+9-n)(2a+9+n)=1939$.
Mà ta có: $1939=\left[ \begin{array}{l}1.1939\\7.277\end{array} \right.$
Ta có 2 trường hợp:
TH1: $$\left \{ {{2a+9-n=1} \atop {2a+9+n=1939}} \right.$$ $⇒n=969$
TH2: $$\left \{ {{2a+9-n=7} \atop {2a+9+n=277}} \right.$$ $⇒n=135$
Vậy $n={969;135}$
