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Question Number 13394 by Tinkutara last updated on 19/May/17

$$\mathrm{Show}\mathrm{that}\mathrm{any}\mathrm{circle}\mathrm{with}\mathrm{centre}(\sqrt{2},\sqrt{3})$$$$\mathrm{cannot}\mathrm{pass}\mathrm{through}\mathrm{more}\mathrm{than}\mathrm{one}$$$$\mathrm{lattice}\mathrm{point}.[\mathrm{Lattice}\mathrm{points}\mathrm{are}\mathrm{points}$$$$\mathrm{in}\mathrm{cartesian}\mathrm{plane},\mathrm{whose}\mathrm{abscissa}\mathrm{and}$$$$\mathrm{ordinate}\mathrm{both}\mathrm{are}\mathrm{integers}.]$$

Answered by mrW1 last updated on 20/May/17

$$circlethroughlatticepoint(i,j)withi,jϵ\mathbb{Z}$$$${(i-\sqrt{2})}^2+{(j-\sqrt{3})}^2=r^2$$$$$$$$foranyotherpoint(x,y)onthiscircle,$$$${(x-\sqrt{2})}^2+{(y-\sqrt{3})}^2=r^2={(i-\sqrt{2})}^2+{(j-\sqrt{3})}^2$$$${(x-\sqrt{2})}^2-{(i-\sqrt{2})}^2+{(y-\sqrt{3})}^2-{(j-\sqrt{3})}^2=0$$$$(x+i-2\sqrt{2})(x-i)+(y+j-2\sqrt{3})(y-j)=0$$$$(x+i)(x-i)-2(x-i)\sqrt{2}+(y+j)(y-j)-2(y-j)\sqrt{3}=0$$$$x^2-i^2+y^2-j^2=2(x-i)\sqrt{2}+2(y-j)\sqrt{3}$$$$$$$$case1:$$$$ifx\ne iandy=j$$$$x^2-i^2=2(x-i)\sqrt{2}$$$$\Rightarrow \sqrt{2}=\frac{x+i}{2}=\frac{integer}{2}$$$$butthisisfalse,since\sqrt{2}isirrational.$$$$$$$$case2:$$$$ifx=iandy\ne j$$$$y^2-j^2=2(y-j)\sqrt{3}$$$$\Rightarrow \sqrt{3}=\frac{y+j}{2}=\frac{integer}{2}$$$$butthisisfalse,since\sqrt{3}isirrational.$$$$$$$$case3:$$$$ifx\ne iandy\ne j$$$${(x^2-i^2+y^2-j^2)}^2=8{(x-i)}^2+12{(y-j)}^2+8(x-i)(y-j)\sqrt{6}$$$$\Rightarrow \sqrt{6}=\frac{{(x^2-i^2+y^2-j^2)}^2-8{(x-i)}^2-12{(y-j)}^2}{8(x-i)(y-j)}$$$$\Rightarrow \sqrt{6}=\frac{integer}{integer}$$$$butthisisfalse,since\sqrt{6}isirrational.$$$$$$$$\Rightarrow xandycannotbeintegerboth,$$$$exceptx=iandy=j.$$

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