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Integer Function Finding (Posted on 2015-10-09) Difficulty: 3 of 5
Find all integer valued functions G:Z→Z such that:
G(1)=G(-1) and:
G(x) + G(y) = G(x+2xy) + G(y-2xy)

No Solution Yet Submitted by K Sengupta    
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Some Thoughts computer guided guess Comment 1 of 1
Below are computer-generated instances of the given formula, with annotations of deductions placed to the right:

g(0) + g(0) = g(0) + g(0)
g(0) + g(0) = g(0) + g(0)
g(0) + g(0) = g(0) + g(0)
g(0) + g(0) = g(0) + g(0)
g(0) + g(-1) = g(0) + g(-1)
g(0) + g(1) = g(0) + g(1)
g(0) + g(-1) = g(0) + g(-1)
g(0) + g(1) = g(0) + g(1)
g(-1) + g(0) = g(-1) + g(0)
g(-1) + g(0) = g(-1) + g(0)
g(1) + g(0) = g(1) + g(0)
g(1) + g(0) = g(1) + g(0)
g(0) + g(-2) = g(0) + g(-2)
g(0) + g(2) = g(0) + g(2)
g(0) + g(-2) = g(0) + g(-2)
g(0) + g(2) = g(0) + g(2)


g(-1) + g(-1) = g(1) + g(-3)   g(-3)=g(1)
g(-1) + g(1) = g(-3) + g(3)    g(3)=g(1)

g(1) + g(-1) = g(-1) + g(1)
g(1) + g(1) = g(3) + g(-1)
g(-2) + g(0) = g(-2) + g(0)
g(-2) + g(0) = g(-2) + g(0)
g(2) + g(0) = g(2) + g(0)
g(2) + g(0) = g(2) + g(0)

g(0) + g(-3) = g(0) + g(-3)    g(0)=g(1)

g(0) + g(3) = g(0) + g(3)
g(0) + g(-3) = g(0) + g(-3)
g(0) + g(3) = g(0) + g(3)

g(-1) + g(-2) = g(3) + g(-6)   g(-6)=g(-2)
g(-1) + g(2) = g(-5) + g(6)    g(1)+g(2)=g(-5)+g(6)
g(1) + g(-2) = g(-3) + g(2)    g(-2) = g(2)
g(1) + g(2) = g(5) + g(-2)     g(5)=g(1)
g(-2) + g(-1) = g(2) + g(-5)   g(-5)=g(1); therefore g(6)=g(2)
g(-2) + g(1) = g(-6) + g(5)    g(-6)=g(2)=g(6) 
g(2) + g(-1) = g(-2) + g(3)
g(2) + g(1) = g(6) + g(-3)

g(-3) + g(0) = g(-3) + g(0)
g(-3) + g(0) = g(-3) + g(0)
g(3) + g(0) = g(3) + g(0)
g(3) + g(0) = g(3) + g(0)
g(0) + g(-4) = g(0) + g(-4)
g(0) + g(4) = g(0) + g(4)
g(0) + g(-4) = g(0) + g(-4)
g(0) + g(4) = g(0) + g(4)

g(-1) + g(-3) = g(5) + g(-9)   g(5)+g(-9) = 2*g(1)
g(-1) + g(3) = g(-7) + g(9)    g(-7)+g(9) = 2*g(1)
g(1) + g(-3) = g(-5) + g(3)    g(-5)=g(1)
g(1) + g(3) = g(7) + g(-3)     g(7)=g(1)
g(-2) + g(-2) = g(6) + g(-10)  g(6)+g(-10)=2*g(2)
g(-2) + g(2) = g(-10) + g(10)  g(10)+g(-10)=2*g(2); therefore g(10)=g(6)
g(2) + g(-2) = g(-6) + g(6)    g(6)+g(-6)=2*g(2)
g(2) + g(2) = g(10) + g(-6)
g(-3) + g(-1) = g(3) + g(-7)   g(-7)=g(1)
g(-3) + g(1) = g(-9) + g(7)
g(3) + g(-1) = g(-3) + g(5)
g(3) + g(1) = g(9) + g(-5)
g(-4) + g(0) = g(-4) + g(0)
g(-4) + g(0) = g(-4) + g(0)
g(4) + g(0) = g(4) + g(0)
g(4) + g(0) = g(4) + g(0)
g(0) + g(-5) = g(0) + g(-5)
g(0) + g(5) = g(0) + g(5)
g(0) + g(-5) = g(0) + g(-5)
g(0) + g(5) = g(0) + g(5)
g(-1) + g(-4) = g(7) + g(-12)
g(-1) + g(4) = g(-9) + g(12)
g(1) + g(-4) = g(-7) + g(4)
g(1) + g(4) = g(9) + g(-4)
g(-2) + g(-3) = g(10) + g(-15)
g(-2) + g(3) = g(-14) + g(15)
g(2) + g(-3) = g(-10) + g(9)
g(2) + g(3) = g(14) + g(-9)
g(-3) + g(-2) = g(9) + g(-14)
g(-3) + g(2) = g(-15) + g(14)
g(3) + g(-2) = g(-9) + g(10)
g(3) + g(2) = g(15) + g(-10)
g(-4) + g(-1) = g(4) + g(-9)
g(-4) + g(1) = g(-12) + g(9)
g(4) + g(-1) = g(-4) + g(7)
g(4) + g(1) = g(12) + g(-7)
g(-5) + g(0) = g(-5) + g(0)
g(-5) + g(0) = g(-5) + g(0)
g(5) + g(0) = g(5) + g(0)
g(5) + g(0) = g(5) + g(0)

It would seem that G(1) and G(2) can be arbitrarily chosen and then all the nonzero (positive and negative) even x have G(x) = G(2) and odd x have G(x)=G(1). G(0) would also be the same as G(1), the only such even case. 

DefDbl A-Z
Dim crlf$


Private Sub Form_Load()
 Form1.Visible = True
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)

 For tot = 0 To 5
   For x0 = 0 To tot
     y0 = tot - x0
     For xf = -1 To 1 Step 2
     For yf = -1 To 1 Step 2
       Text1.Text = Text1.Text & "g(" & x0 * xf & ") + "
       Text1.Text = Text1.Text & "g(" & y0 * yf & ") = "
       Text1.Text = Text1.Text & "g(" & x0 * xf + 2 * x0 * xf * y0 * yf & ") + "
       Text1.Text = Text1.Text & "g(" & y0 * yf - 2 * x0 * xf * y0 * yf & ")"
       Text1.Text = Text1.Text & crlf
     Next
     Next
   Next x0
 Next tot

 Text1.Text = Text1.Text & crlf & " done"
  
End Sub


  Posted by Charlie on 2015-10-09 11:22:43
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