```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
```