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Question: Find, with explanation, all integers x, y with x(y + 1)^2 = 243y. Solution: 1. Take x = 24, y = 8 Left Hand Side = 24 ( 8 1 ) 2 Right Hand Side = 243 * 8 24 * 81 1944 1944 Hence 2(y+1)^2 = 243y is true for x = 24, y = 8 2. Take x = 54, y = 2 Left Hand Side = 54 ( 2 1 ) 2 54 * 9 486 Right Hand Side = 243 * 2 486 Hence 2(y+1)^2 = 243y is true for x = 54, y = 2 We can use mathematical reasoning to exclude the possibility of another solution as below: Expanding the equation and building a quadratic in y:

*x*

(

*y*

2 2

*y*

1 ) 243

*y*

or or

*xy*

2

*xy*

2 2

*xy*

*x*

243

*y*

( 2

*x*

243 )

*y*

*x*

0 0 This is a quadratic of the type Which has real roots only if Or ( 2

*x*

243 ) 2 4 *

*x*

*

*x b*

2

*ay*

2

*by*

*c*

4

*ac*

0 Or Or 4

*x*

2 59049 59049 972

*x*

972

*x*

4

*x*

2 Or

*x*

59049 60 .

75 972 Since by trial and error, we get two solutions with x = 24, and 54, and from x values 55 to 60, we get no other solution, we conclude that these are the only two solutions possible. The following is a Visual Basic macro which gives the solution for all integers in the range 1 to 1000. Private Sub Calculate() For x = 1 To 1000 For y = 1 To 1000 If (x * (y + 1) ^ 2 = 243 * y) Then ActiveCell.Value = x ActiveCell.Offset(0, 1).Value = y ActiveCell.Offset(1, 0).Activate End If Next y Next x End Sub