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4:
(x+1)(y-2)=5
=>\(\left(x+1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)
`= x(y+2) + y+2 = 3`.
`=> (x+1)(y+2) = 3`
`=> x + 1 in Ư(3)`
`=> x + 1 = 1 => x = 0`
`y + 2 = 3 => y = 1`.
`=> x +1 = -1 => x = -2`
`y + 2 = -3 => y = -5`
`=> x + 1 = 3 => x = 2`
`y + 2 = 1 => y = -1`
`=> x + 1 = -3 => x = -4`
`y + 2 = -1 => y = -3`.
`= x(y+2) + y+2 = 3`.
`=> (x+1)(y+2) = 3`
`=> x + 1 in Ư(3)`
`=> x + 1 = 1 => x = 0`
`y + 2 = 3 => y = 1`.
`=> x +1 = -1 => x = -2`
`y + 2 = -3 => y = -5`
`=> x + 1 = 3 => x = 2`
`y + 2 = 1 => y = -1`
`=> x + 1 = -3 => x = -4`
`y + 2 = -1 => y = -3`.
\(x^2+x+1=y^2\)
\(\Leftrightarrow4x^2+4x+4=4y^2\)
\(\Leftrightarrow4y^2-\left(4x^2+4x+1\right)=3\)
\(\Leftrightarrow\left(2y-2x-1\right)\left(2y+2x+1\right)=3\)
Do \(x,y\inℤ\)nên \(2y-2x-1,2y+2x+1\)là ước của \(3\).
Ta có bảng sau:
2y-2x-1 | 1 | 3 | -1 | -3 |
2y+2x+1 | 3 | 1 | -3 | -1 |
x | 0 | -1 | -1 | 0 |
y | 1 | 1 | -1 | -1 |
\(x^2-xy+y+2=0\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2-\left(\dfrac{y}{2}-1\right)^2+3=0\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2-\left(\dfrac{y}{2}-1\right)^2=1-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{y}{2}\right)^2=1\\\left(\dfrac{y}{2}-1\right)^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x-y=2\\2x-y=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=6\\y=-2\end{matrix}\right.\end{matrix}\right.\)
với y=6 \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
với y=-2 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
vậy S=\(\left\{\left(4;6\right);\left(2;6\right);\left(0;-2\right);\left(-2;-2\right)\right\}\)