# Find, with explanation, all positive integers x, y, z with 2^x + 3^y = z^2 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