Question

TÌm nghiệm nguyên của phương trình: \(x^2+2x-10=y^2\)

Answer 1

\(\Leftrightarrow x^2+2x+1=y^2+11\)

\(\Leftrightarrow\left(x+1\right)^2-y^2=11\)

\(\Leftrightarrow\left(x+1-y\right)\left(x+1+y\right)=11\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1-y=-1\\x+1+y=-11\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-y=-11\\x+1+y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-y=1\\x+1+y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-y=11\\x+1+y=1\end{matrix}\right.\end{matrix}\right.\)

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