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 24 * 81 1944 Right Hand Side = 243 * 8 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 xy 2 2 xy x 243 y 0 or xy 2 (2 x 243) y x 0 This is a quadratic of the type ay 2 by c 0 Which has real roots only if b 2 4ac Or (2 x 243) 2 4 * x * x Or 4 x 2 59049 972 x 4 x 2 Or 59049 972x 59049 60.75 Or x 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

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