Xét \(y\left(y+3\right)\left(y+1\right)\left(y+2\right)=\left(y^2+3y\right)\left(y^2+3y+2\right)\)
\(=\left(y^2+3y\right)^2+2\left(y^2+3y\right)+1-1\)
\(=\left(y^2+3y+1\right)^2-1\)
\(\Rightarrow\left(x-2015\right)^2=\left(y^2+3y+1\right)^2-1\)
Để cho gọn ta đặt \(\left\{{}\begin{matrix}x-2015=a\\y^2+3y+1=b\end{matrix}\right.\) với \(a;b\in Z\)
\(\Rightarrow a^2=b^2-1\Rightarrow b^2-a^2=1\)
\(\Rightarrow\left(b-a\right)\left(b+a\right)=1\)
TH1: \(\left\{{}\begin{matrix}b-a=1\\b+a=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0\\b=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2015=0\\y^2+3y+1=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2015\\y=\left\{0;-3\right\}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}b-a=-1\\b+a=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0\\b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2015\\y=\left\{-1;-2\right\}\end{matrix}\right.\)
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