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Equal Power Sum Sets (Posted on 2017-10-11) Difficulty: 3 of 5
41+51+121=21+91+101
42+52+122=22+92+102

Are equal power sum sets (EPSS) for second power, but the sum 4+5+12 is not the smallest. Find the smallest with all distinct positive integers.

Find EPSS of four numbers each (8 distinct positive integers) where the first, second and third powers have equal sums.

Keep going, if possible, to find higher order EPSS where fourth, fifth powers, etc. have equal sums.

Note: The highest EPSS I have is fifth power.

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Solution computer solution Comment 1 of 1
Part 1:

1+5+6 = 2+3+7 and
1^2 + 5^2 + 6^2 = 2^2 + 3^2 + 7^2 

Other pairs of sets are shown (those using no number higher than 12):

Duplicates occur as no check was done for reversing LHS and RHS.

1 5 6     2 3 7
1 6 8     2 4 9
1 7 10     2 5 11
1 8 9     3 4 11
2 3 7     1 5 6
2 4 9     1 6 8
2 5 11     1 7 10
2 6 7     3 4 8
2 7 9     3 5 10
2 8 11     3 6 12
2 9 10     4 5 12
3 4 8     2 6 7
3 4 11     1 8 9
3 5 10     2 7 9
3 6 12     2 8 11
3 7 8     4 5 9
3 8 10     4 6 11
4 5 9     3 7 8
4 5 12     2 9 10
4 6 11     3 8 10
4 8 9     5 6 10
4 9 11     5 7 12
5 6 10     4 8 9
5 7 12     4 9 11
5 9 10     6 7 11
6 7 11     5 9 10
6 10 11     7 8 12
7 8 12     6 10 11

DefDbl A-Z
Dim crlf$, used(12)


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 
 For a = 1 To 12
   used(a) = 1
 For b = a + 1 To 12
   used(b) = 1
 For c = b + 1 To 12
   used(c) = 1
 For x = 1 To 12
  If used(x) = 0 Then
   used(x) = 1
 For y = x + 1 To 12
  If used(y) = 0 Then
   used(y) = 1
 For z = y + 1 To 12
  If used(z) = 0 Then
   used(z) = 1
   
   If a + b + c = x + y + z Then
     If a * a + b * b + c * c = x * x + y * y + z * z Then
       Text1.Text = Text1.Text & a & Str(b) & Str(c) & "    " & Str(x) & Str(y) & Str(z) & crlf
     End If
   End If
   
   used(z) = 0
  End If
  DoEvents
 Next z
   used(y) = 0
  End If
 Next y
   used(x) = 0
  End If
 Next x
   used(c) = 0
 Next c
   used(b) = 0
 Next b
   used(a) = 0
 Next a
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub


Part 2:

The below is, again, really only one solution with left and right sides of the equation reversed. The first row in each case shows the first power; the second row, the second power; and the third row, the third power.

1 5 8 12     2 3 10 11
1 25 64 144      4 9 100 121
1 125 512 1728        8 27 1000 1331

2 3 10 11     1 5 8 12
4 9 100 121      1 25 64 144
8 27 1000 1331        1 125 512 1728


DefDbl A-Z
Dim crlf$, used(12)


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 
 For a = 1 To 12
   used(a) = 1
 For b = a + 1 To 12
   used(b) = 1
 For c = b + 1 To 12
   used(c) = 1
 For d = c + 1 To 12
   used(d) = 1
 For x = 1 To 12
  If used(x) = 0 Then
   used(x) = 1
 For y = x + 1 To 12
  If used(y) = 0 Then
   used(y) = 1
 For z = y + 1 To 12
  If used(z) = 0 Then
   used(z) = 1
 For w = z + 1 To 12
  If used(w) = 0 Then
   used(w) = 1
   
   If a + b + c + d = x + y + z + w Then
     If a * a + b * b + c * c + d * d = x * x + y * y + z * z + w * w Then
     If a * a * a + b * b * b + c * c * c + d * d * d = x * x * x + y * y * y + z * z * z + w * w * w Then
       Text1.Text = Text1.Text & a & Str(b) & Str(c) & Str(d) & "    " & Str(x) & Str(y) & Str(z) & Str(w) & crlf
       Text1.Text = Text1.Text & a ^ 2 & Str(b ^ 2) & Str(c ^ 2) & Str(d ^ 2) & "     " & Str(x ^ 2) & Str(y ^ 2) & Str(z ^ 2) & Str(w ^ 2) & crlf
       Text1.Text = Text1.Text & a ^ 3 & Str(b ^ 3) & Str(c ^ 3) & Str(d ^ 3) & "       " & Str(x ^ 3) & Str(y ^ 3) & Str(z ^ 3) & Str(w ^ 3) & crlf
       Text1.Text = Text1.Text & crlf
     End If
     End If
   End If
   
   used(w) = 0
  End If
  DoEvents
 Next w
   used(z) = 0
  End If
 Next z
   used(y) = 0
  End If
 Next y
   used(x) = 0
  End If
 Next x
   used(d) = 0
 Next d
   used(c) = 0
 Next c
   used(b) = 0
 Next b
   used(a) = 0
 Next a
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub


  Posted by Charlie on 2017-10-11 15:13:30
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