\(\Leftrightarrow\left(2019x+2020\right)^2=\left(y+1\right)\left(y^2-y+1\right)\)
Gọi \(d=ƯCLN\left(y+1;y^2-y+1\right)\)
\(\Rightarrow y\left(y+1\right)-\left(y^2-y+1\right)⋮d\)
\(\Rightarrow2y-1⋮d\)
\(\Rightarrow2\left(y+1\right)-\left(2y-1\right)⋮d\)
\(\Rightarrow3⋮d\Rightarrow\left[{}\begin{matrix}d=1\\d=3\end{matrix}\right.\)
- Với \(d=1\Rightarrow y+1\) và \(y^2-y+1\) nguyên tố cùng nhau
Mà vế trái là SCP \(\Rightarrow\left\{{}\begin{matrix}y+1=a^2\\y^2-y+1=b^2\end{matrix}\right.\)
Nhưng do y nguyên dương \(\Rightarrow y\ge1\Rightarrow-y+1\le0\)
\(\Rightarrow y^2-y+1\le y^2\Rightarrow\left(y-1\right)^2< y^2-y+1\le y^2\)
\(\Rightarrow y^2-y+1=y^2\Rightarrow y=1\)
\(\Rightarrow y+1=2\) ko phải SCP (loại)
- Với \(d=3\Rightarrow VP⋮3\)
Mà \(\left\{{}\begin{matrix}2019⋮3\\2020⋮̸3\end{matrix}\right.\) \(\Rightarrow2019x+2020⋮3̸\Rightarrow VT⋮̸3\)
Vậy pt vô nghiệm
