Question

[Nguyễn Bùi Đại Hiệp]

Tìm các cặp số nguyên (x,y) thỏa mãn 2x²+3y²+4x=19

Answer 1

[Nguyễn Thành Trương]

Ta có: \(2x^2+3y^2+4x=19\)

\(\Leftrightarrow\) \(2x^2+4x=19-3y^2\)

\(\Leftrightarrow\) \(2x^2+4x+2=21-3y^2\)

\(\Leftrightarrow\) \(2\left(x+1\right)^2=3\left(7-y^2\right)\) \(\left(\text{*}\right)\)

Vì \(2\left(x+1\right)^2\) chia hết cho \(2\) nên \(3\left(7-y^2\right)\) chia hết cho \(2\), hay \(7-y^2\) chia hết cho \(2\) , hay \(y^2\) lẻ \(\left(1\right)\)

Lại có: \(7-y^2\ge0\) (do \(\left(x+1\right)^2\ge0\)) nên \(y^2\le7\) (với \(y\in Z\) ), tức là \(y^2\in\left\{1;4\right\}\) \(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) , suy ra \(y^2=1\) \(\Rightarrow\) \(y\in\left\{-1;1\right\}\)

Khi đó, phương trình \(\left(\text{*}\right)\) sẽ có dạng \(2\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(x+1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy, các cặp nghiệm nguyên phải tìm: \(\left(x;y\right)=\left\{\left(2;1\right),\left(2;-1\right),\left(-4;1\right),\left(-4;-1\right)\right\}\) (thỏa mãn \(x,y\in Z\) )

Answer 2

[Nguyễn Việt Lâm]

\(2\left(x+1\right)^2+3y^2=21\Rightarrow y^2=7-\frac{2\left(x+1\right)^2}{3}\le7\)

\(\Rightarrow y^2=\left\{0;1;4\right\}\)

- \(y^2=0\Rightarrow y=0\Rightarrow\left(x+1\right)^2=\frac{21}{2}\) (ko có x nguyên thỏa mãn)

- \(y^2=1\Rightarrow y=\pm1\Rightarrow\left(x+1\right)^2=\frac{21-3y^2}{2}=9\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

- \(y^2=4\Rightarrow\left(x+1\right)^2=\frac{21-3y^2}{2}=\frac{9}{2}\) (ko có x nguyên thỏa mãn)

Vậy \(\left(x;y\right)=\left(2;1\right);\left(2;-1\right);\left(-4;1\right);\left(-4;-1\right)\)