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Let $f(x,y) = 4x^2 -3xy -2y + 1 $ . Find $(x , y)$ so that $x , y \in \mathbb{Z} $ and $f(x,y) = 0 $ . I've tried many numbers and got some answers but it's not the solution !

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    $\begingroup$ $3x+2$ will divide $$4x^2+1$$ iff $3x+2$ will divide $$3(4x^2+1)=4x(3x+2)+3-8x$$ $3x+2$ will divide $3-8x$ iff $3x+2$ will divide $$3(3-8x)=25-8(3x+2)$$ so, $3x+2$ needs to divide $25$ $\endgroup$ – lab bhattacharjee Dec 26 '17 at 12:17
  • $\begingroup$ Very nice , Thanks ! $\endgroup$ – S.H.W Dec 26 '17 at 12:40
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$$f(x,y)=4x^2 -3xy -2y + 1$$

$$y=\frac{4x^2+1}{3x+2},~~ 3x+2\ne 0$$

$$3y=\frac{12x^2+3}{3x+2}=4x-\frac{8x-3}{3x+2}=4x-2-\frac{2x-7}{3x+2}$$

$$3(-3y+4x-2)=\frac{6x-21}{3x+2}=2+\frac{-25}{3x+2}$$

$$9y-12x+8=\frac{25}{3x+2}$$

$$(9y-12x+8)(3x+2)=25$$

$$k(3x+2)=25$$

$$3x+2=-25,-5,-1,1,5,25$$

Try

$$ \begin{align} & 3x+2=-25,~~~k=-1,~~~~x=-9,~~~~y=-13 \\ & 3x+2=-5,~~~~~k=-5,~~~~x=\text{bad}\\ &3x+2=-1,~~~~k=-25,~~~~x=-1,~~~~y=-5\\ &3x+2=+1,~~~~k=25,~~~~~~x=\text{bad}\\ &3x+2=+5,~~~~~k=+5,~~~~x=1,~~~~y=1\\ &3x+2=25,~~~~~~k=+1,~~~~x=\text{bad}\\ \end{align} $$

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$$f(x,y)=0$$ $$\Rightarrow 4x^2-3xy-2y+1=0$$ $$\Rightarrow 4x^2-3xy-y^2+y^2-2y+1=0$$ $$\Rightarrow 4x^2-4xy+xy-y^2+(y-1)^2=0$$ $$\Rightarrow (4x-y)(x-y)+(y-1)^2=0$$ $$\Rightarrow (4x-y)(x-y)+(y-1)^2=0$$

Since $(y-1)^2 \ge 0$, so we can conclude that $(4x-y)(x-y)$ must be negative for $f(x,y)=0$ to hold.

Which means, either $4x-y \le 0$ and $x+y \ge 0$; OR
$x+y \le 0$ and $4x-y \ge 0$.

See if this helps.

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$\left(-9,-13\right)$,$\left(-1,-5\right)$,$\left(1,1\right)$. Three pairs.

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