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20/12/2017 at 20:35##### Nguyễn Hưng Phát

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23/12/2017 at 14:42-
FC Alan Walker 16/03/2018 at 13:31
We have: \(y^2=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

\(\Leftrightarrow y^2+1=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(\Leftrightarrow y^2+1=\left(x^2+3x+1\right)^2\)

\(\Rightarrow y=0\) (Because \(y^2\) and \(\left(x^2+3x+1\right)^2\) is two consecutive square numbers)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)

So x=0; x=-1; x=-2; x=-3.

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