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Question Number 19413 by Tinkutara last updated on 10/Aug/17

$$\mathrm{How}\mathrm{many}\mathrm{integer}\mathrm{pairs}(x,y)\mathrm{satisfy}$$$$x^2+4y^2-2xy-2x-4y-8=0?$$

Commented by mrW1 last updated on 11/Aug/17

$$(0,-1)$$$$(-2,0)$$$$(4,0)$$$$(0,2)$$$$(6,2)$$$$(4,3)$$$$\mathrm{is}\mathrm{this}\mathrm{correct}?$$

Answered by mrW1 last updated on 11/Aug/17

$$x^2-2(y+1)x-4(2+y-{\mathrm{y}}^2)=0$$$$\Rightarrow \mathrm{x}=\frac{2(\mathrm{y}+1)\pm \sqrt{4{(\mathrm{y}+1)}^2+16(2+\mathrm{y}-{\mathrm{y}}^2)}}{2}$$$$\mathrm{x}=\mathrm{y}+1\pm \sqrt{{\mathrm{y}}^2+2\mathrm{y}+1+8+4\mathrm{y}-{4\mathrm{y}}^2}$$$$\mathrm{x}=\mathrm{y}+1\pm \sqrt{3(3+2\mathrm{y}-{\mathrm{y}}^2)}$$$$\mathrm{x}=\mathrm{y}+1\pm \sqrt{3[4-{(\mathrm{y}-1)}^2]}$$$$4-{(\mathrm{y}-1)}^2⩾0$$$${(\mathrm{y}-1)}^2⩽4$$$$\mathrm{since}\mathrm{x}\mathrm{is}\mathrm{integer},4-{(\mathrm{y}-1)}^2=0\mathrm{or}3$$$$\Rightarrow {(\mathrm{y}-1)}^2=4\mathrm{or}1$$$$\Rightarrow \mathrm{y}-1=\pm 2\mathrm{or}\pm 1$$$$\Rightarrow \mathrm{y}=1\pm 2\mathrm{or}1\pm 1$$$$\Rightarrow \mathrm{y}=3,-1,2,0$$$$\Rightarrow \mathrm{x}=4,0,3\pm 3,1\pm 3$$$$\Rightarrow (\mathrm{x},\mathrm{y})=(4,3)/(0,-1)/(6,2)/(0,2)/(4,0)/(-2,0)$$

Commented by Tinkutara last updated on 11/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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