Question Number 103093 by mr W last updated on 12/Jul/20

solve for x,y∈N  (1/x)+(1/y)=(3/(202))

Answered by 1549442205 last updated on 13/Jul/20

We have (1/x)+(1/y)=(3/(202))⇔((x+y)/(xy))=(3/(202))  ⇔3xy=202(x+y)⇔9xy=3.202(x+y)  ⇔(3x−202)(3y−202)=202^2   ⇒3x−202∣202^2 =101^2 ×2^2   ⇒3x−202∈{1,2,4,101,202,404,10201,20402,40804}  The case 1,4,202,10201,40804 is reject  because when  plus to 202 we obtain  a number which isn′t divisible by 3.  We have following tables:   determinant (((3x−202),2,(101),(404),(20402)),((3y−202),(202×101),(101×4),(101),2),(x,(68),(101),(202),(6868)),(y,(6868),(202),(101),(68)))  Thus,the set of  solutions as  {(68;6868);(101;202);(202;101);(6868;68)}

Commented bymr W last updated on 13/Jul/20

thanks sir!

Answered by Rasheed.Sindhi last updated on 13/Jul/20

((x+y)/(xy))=(3/(202))  x+y=3k ∧ xy=202k          xy=1×2×101×k  possible values for x(or y):  x=1,2,101,202,k,2k,101k,202k  y=((202k)/x)=202k,101k,2k,k,202,101,2,1  y=3k−x=3k−1,3k−2,3k−101,3k−202,   determinant ((x,(          y),(       y),(       k),x,y),(,(=(202k)/x),(=3k−x),,,),(1,(202k),(3k−1),(−1/199),1,−),(2,(101k),(3k−2),(−1/49),2,−),((101),(2k),(3k−101),(101),(101),(202)),((202),k,(3k−202),(101),(202),(101)),(k,(202),(2k),(101),(101),(202)),((2k),(101),k,(101),(202),(101)),((101k),2,(−98k),(−1/49),−,2),((202k),1,(−199k),(−1/199),−,1))               x=k  ((k+y)/(ky))=(3/(202))  y=((202k)/(3k−202))  3k−202=1,2,101,202,k,2k,101k,202k  3k−202=k,2k,101k,202k (in general)  k=101,202,((−101)/(49)),((−202)/(199))  y=202,101  x=101,202  (_★ ^★ )  x=2k  ((2k+y)/(2ky))=(3/(202))  404k+202y=6ky  y=((202k)/(3k−101))  3k−101=1,2,101,202,k,2k,101k,202k  3k−101=k,2k,101k,202k(in general)  k=((101)/2),101,((−101)/(98)),((−101)/(199))  for k=101  y=202  x=101  3k−101=1⇒k=34⇒y=6868  3k−101=2k⇒k=101⇒y=101  3k−101=101k⇒y∉N  3k−101=202k⇒y∉N    y=101k  ....  y=202k  .......

Commented byDwaipayan Shikari last updated on 12/Jul/20

There is an another solution( 68,6868) or (6868,68)  ∈N

Commented bymr W last updated on 12/Jul/20

how did you get it and how are you  sure that no other solutions exist?

Answered by nimnim last updated on 12/Jul/20

The simplest way to get a   solution would be:  (1/x)+(1/y)=(3/(202))=((2+1)/(202))=(1/(101))+(1/(202))  ∴(x,y)=(101,202) or (202,101)

Commented bymr W last updated on 12/Jul/20

correct, thanks sir!  but how are you sure if there are no  other solutions?

Answered by floor(10²Eta[1]) last updated on 12/Jul/20

(1/x)=(3/(202))−(1/y)=((3y−202)/(202y))  (1/y)=((3x−202)/(202x))  ⇒(1/x)×(1/y)=(((3y−202)(3x−202))/(202^2 xy))  ⇒(3y−202)(3x−202)=202^2 =2^2 .101^2   3y−202≡2(mod 3) and 3x−202≡2(mod 3)  so a good strategy is look for factor pairs  of this right hand side of the equation where  each part of the factor pair is 2 mod 3.  (1).(2^2 .101^2 )≢2(mod 3)  (2).(2.101^2 )≡2(mod 3)  (2^2 ).(101^2 )≢2(mod 3)  (2^2 .101).(101)≡2(mod 3)  so now we will assigne each part of this  factor pairs to each one of these binomials  (3y−202 and 3x−202) and solve for x or y  1 case: 3y−202=2⇒y=68  3x−202=2.101^2 ⇒x=6868  2 case: 3y−202=2^2 .101⇒y=202  3x−202=101⇒x=101  ⇒all solutions are   (x, y)={(6868; 68), (68; 6868),   (101; 202), (202; 101)}.

Commented bymr W last updated on 12/Jul/20

thanks alot sir!

Answered by Rasheed.Sindhi last updated on 14/Jul/20

An Easy ∧ Short approach  (1/y)=(3/(202))−(1/x)=((3x−202)/(202x))     y=((202x)/(3x−202))⇒3x−202 ∣ 202x    possible values for 3x−202   _(−)   3x−202=1,2,101,202,x,2x,101x,202x   determinant (((3x−202=),(                x),(   y)),((         1),(((1+202)/3)=((203)/3)∉N),( −)),((         2),(((2+202)/3)=68),(6868)),((      101),(((101+202)/3)=101),( 202)),((      202),(((202+202)/3)=((404)/3)∉N),(  −)),((        x),(2x=202⇒x=101),( 202)),((      2x),(x=202),( 101)),((    101x),(98x=−202⇒x∉N),(  −)),((    202x),(199x=−202⇒x∉N),(  −)))

Commented bymr W last updated on 17/Jul/20

great! thanks alot!