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 =

24 * 81

1944
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
Right Hand Side =

1 ) 2 
54 * 9
243 * 2

486

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 ay
2  by
 c

Which has real roots only if b
2 
4 ac
Or ( 2 x

243 )
2 
4 * x * x
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