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

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

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

Answered by mrW1 last updated on 20/May/17

circle through lattice point (i,j) with i,jεZ (i−(√2))^2 +(j−(√3))^2 =r^2 for any other point (x,y) on this circle, (x−(√2))^2 +(y−(√3))^2 =r^2 =(i−(√2))^2 +(j−(√3))^2 (x−(√2))^2 −(i−(√2))^2 +(y−(√3))^2 −(j−(√3))^2 =0 (x+i−2(√2))(x−i)+(y+j−2(√3))(y−j)=0 (x+i)(x−i)−2(x−i)(√2)+(y+j)(y−j)−2(y−j)(√3)=0 x^2 −i^2 +y^2 −j^2 =2(x−i)(√2)+2(y−j)(√3) case 1: if x≠i and y=j x^2 −i^2 =2(x−i)(√2) ⇒(√2)=((x+i)/2)=((integer)/2) but this is false, since (√2) is irrational. case 2: if x=i and y≠j y^2 −j^2 =2(y−j)(√3) ⇒(√3)=((y+j)/2)=((integer)/2) but this is false, since (√3) is irrational. case 3: if x≠i and y≠j (x^2 −i^2 +y^2 −j^2 )^2 =8(x−i)^2 +12(y−j)^2 +8(x−i)(y−j)(√6) ⇒(√6)=(((x^2 −i^2 +y^2 −j^2 )^2 −8(x−i)^2 −12(y−j)^2 )/(8(x−i)(y−j))) ⇒(√6)=((integer)/(integer)) but this is false, since (√6) is irrational. ⇒x and y can not be integer both, except x=i and y=j.

$${circle}\:{through}\:{lattice}\:{point}\:\left({i},{j}\right)\:{with}\:{i},{j}\epsilon\mathbb{Z} \\ $$$$\left({i}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left({j}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$ \\ $$$${for}\:{any}\:{other}\:{point}\:\left({x},{y}\right)\:{on}\:{this}\:{circle}, \\ $$$$\left({x}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} =\left({i}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left({j}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\left({x}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\left({i}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left({j}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}+{i}−\mathrm{2}\sqrt{\mathrm{2}}\right)\left({x}−{i}\right)+\left({y}+{j}−\mathrm{2}\sqrt{\mathrm{3}}\right)\left({y}−{j}\right)=\mathrm{0} \\ $$$$\left({x}+{i}\right)\left({x}−{i}\right)−\mathrm{2}\left({x}−{i}\right)\sqrt{\mathrm{2}}+\left({y}+{j}\right)\left({y}−{j}\right)−\mathrm{2}\left({y}−{j}\right)\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{i}^{\mathrm{2}} +{y}^{\mathrm{2}} −{j}^{\mathrm{2}} =\mathrm{2}\left({x}−{i}\right)\sqrt{\mathrm{2}}+\mathrm{2}\left({y}−{j}\right)\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${case}\:\mathrm{1}: \\ $$$${if}\:{x}\neq{i}\:{and}\:{y}={j} \\ $$$${x}^{\mathrm{2}} −{i}^{\mathrm{2}} =\mathrm{2}\left({x}−{i}\right)\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}=\frac{{x}+{i}}{\mathrm{2}}=\frac{{integer}}{\mathrm{2}} \\ $$$${but}\:{this}\:{is}\:{false},\:{since}\:\sqrt{\mathrm{2}}\:{is}\:{irrational}. \\ $$$$ \\ $$$${case}\:\mathrm{2}: \\ $$$${if}\:{x}={i}\:{and}\:{y}\neq{j} \\ $$$${y}^{\mathrm{2}} −{j}^{\mathrm{2}} =\mathrm{2}\left({y}−{j}\right)\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}=\frac{{y}+{j}}{\mathrm{2}}=\frac{{integer}}{\mathrm{2}} \\ $$$${but}\:{this}\:{is}\:{false},\:{since}\:\sqrt{\mathrm{3}}\:{is}\:{irrational}. \\ $$$$ \\ $$$${case}\:\mathrm{3}: \\ $$$${if}\:{x}\neq{i}\:{and}\:{y}\neq{j} \\ $$$$\left({x}^{\mathrm{2}} −{i}^{\mathrm{2}} +{y}^{\mathrm{2}} −{j}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{8}\left({x}−{i}\right)^{\mathrm{2}} +\mathrm{12}\left({y}−{j}\right)^{\mathrm{2}} +\mathrm{8}\left({x}−{i}\right)\left({y}−{j}\right)\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\sqrt{\mathrm{6}}=\frac{\left({x}^{\mathrm{2}} −{i}^{\mathrm{2}} +{y}^{\mathrm{2}} −{j}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{8}\left({x}−{i}\right)^{\mathrm{2}} −\mathrm{12}\left({y}−{j}\right)^{\mathrm{2}} }{\mathrm{8}\left({x}−{i}\right)\left({y}−{j}\right)}\: \\ $$$$\Rightarrow\sqrt{\mathrm{6}}=\frac{{integer}}{{integer}} \\ $$$${but}\:{this}\:{is}\:{false},\:{since}\:\sqrt{\mathrm{6}}\:{is}\:{irrational}. \\ $$$$ \\ $$$$\Rightarrow{x}\:{and}\:{y}\:{can}\:{not}\:{be}\:{integer}\:{both}, \\ $$$${except}\:{x}={i}\:{and}\:{y}={j}. \\ $$

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