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Non-zero integers (Posted on 2018-06-20) Difficulty: 3 of 5
Find all solutions of
(x+y^2)*(x^2+y)=(x-y)^3 .

x and y have to be non-zero integers.

No Solution Yet Submitted by Ady TZIDON    
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Solution this seems the solution | Comment 1 of 3
(x+y^2)*(x^2+y)=(x-y)^3 

x^3 + xy + (xy)^2 + y^3 = x^3 - 3 x^2 y + 3 x y^2 - y^3

xy + (xy)^2 + y^3 = - 3 x^2 y + 3 x y^2 - y^3

Asking Wolfram Alpha to solve this gives:

If y = -3, x = -9/5

If y <> -3:

x = -(-3 y + sqrt(-(y + 1)^2 (8 y - 1)) + 1)/(2 y + 6) 

or

x = (3 y + sqrt(-(y + 1)^2 (8 y - 1)) - 1)/(2 (y + 3)) 

A map of the continuous curve is at:


but of course we want integers.

So then

 For y = -100000 To 100000
   If y <> -3 Then
    If (-(y + 1) ^ 2 * (8 * y - 1)) >= 0 Then
     x = -(-3 * y + Sqr(-(y + 1) ^ 2 * (8 * y - 1)) + 1) / (2 * y + 6)
     xr = Int(x + 0.5)
     If Abs(x - xr) < 0.00000001 Then Text1.Text = Text1.Text & y & "   " & x & crlf
     
     x = (3 * y + Sqr(-(y + 1) ^ 2 * (8 * y - 1)) - 1) / (2 * (y + 3))
     xr = Int(x + 0.5)
     If Abs(x - xr) < 0.00000001 Then Text1.Text = Text1.Text & y & "   " & x & crlf
    End If
   End If
   DoEvents
 Next
 
 finds
 
  y   x 
-21   9
-10   8
 -6   9
 -1  -1
 -1  -1

  0   0
  
Of course we don't count the last set, as the puzzle specifically excludes zeros. (-1,-1) appears to be a double solution.

  Posted by Charlie on 2018-06-20 13:22:51
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