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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.]
Showthatanycirclewithcentre(2,3)cannotpassthroughmorethanonelatticepoint.[Latticepointsarepointsincartesianplane,whoseabscissaandordinatebothareintegers.]
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.
circlethroughlatticepoint(i,j)withi,jϵZ(i2)2+(j3)2=r2foranyotherpoint(x,y)onthiscircle,(x2)2+(y3)2=r2=(i2)2+(j3)2(x2)2(i2)2+(y3)2(j3)2=0(x+i22)(xi)+(y+j23)(yj)=0(x+i)(xi)2(xi)2+(y+j)(yj)2(yj)3=0x2i2+y2j2=2(xi)2+2(yj)3case1:ifxiandy=jx2i2=2(xi)22=x+i2=integer2butthisisfalse,since2isirrational.case2:ifx=iandyjy2j2=2(yj)33=y+j2=integer2butthisisfalse,since3isirrational.case3:ifxiandyj(x2i2+y2j2)2=8(xi)2+12(yj)2+8(xi)(yj)66=(x2i2+y2j2)28(xi)212(yj)28(xi)(yj)6=integerintegerbutthisisfalse,since6isirrational.xandycannotbeintegerboth,exceptx=iandy=j.

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