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Arithmetic and Power Puzzle (Posted on 2015-09-02) Difficulty: 3 of 5
P,Q,R and S are four positive integers in arithmetic sequence, with P < Q < R < S.

Find all possible solutions such that:

P3 + Q3 + R3 + S3 is a perfect square.

No Solution Yet Submitted by K Sengupta    
Rating: 1.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts computer exploration | Comment 1 of 2
For brevity of space and time, p was limited to no more than 10,000; difference no more than 30,000 and s no more than 100,000.

1 2 3 4
1 33 65 97
4 43 82 121
9 43 77 111
27 49 71 93
39 53 67 81
93 191 289 387
147 1010 1873 2736
210 239 268 297
223 676 1129 1582
358 4801 9244 13687
441 1828 3215 4602
478 1041 1604 2167
487 764 1041 1318
555 914 1273 1632
627 16850 33073 49296
714 1607 2500 3393
795 1634 2473 3312
858 1247 1636 2025
873 6532 12191 17850
982 1369 1756 2143
1122 1895 2668 3441
1156 1787 2418 3049
1204 1283 1362 1441
1293 18071 34849 51627
1488 2017 2546 3075
1582 1929 2276 2623
1593 3340 5087 6834
1684 3643 5602 7561
1687 17564 33441 49318
1801 3081 4361 5641
1849 3977 6105 8233
1929 2083 2237 2391
1945 4601 7257 9913
2017 2993 3969 4945
2226 12191 22156 32121
2304 2617 2930 3243
2422 9689 16956 24223
2473 7897 13321 18745
2902 4649 6396 8143
3265 4081 4897 5713
3265 13601 23937 34273
4198 4481 4764 5047
4276 5107 5938 6769
4543 9916 15289 20662
5187 6770 8353 9936
5236 9227 13218 17209
5302 16809 28316 39823
5472 9745 14018 18291
5527 10244 14961 19678
5889 18604 31319 44034
6121 6722 7323 7924
6699 6722 6745 6768
7041 11347 15653 19959
7153 13825 20497 27169
7437 10319 13201 16083
7465 11929 16393 20857
8067 8449 8831 9213
8163 34081 59999 85917
8497 22785 37073 51361
9009 9892 10775 11658
9262 10569 11876 13183
9417 9700 9983 10266
9433 12905 16377 19849
9546 11831 14116 16401
9867 14738 19609 24480

Each represents a family where you can multiply all four by any perfect square to get a new set.

It doesn't look as if there's an end to such families.


DefDbl A-Z
Dim crlf$


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

 For p = 1 To 10000
  p3 = p * p * p
  DoEvents
  For diff = 1 To 30000
   If gcd(p, diff) = 1 Then
    s = p + 3 * diff
    If s > 100000 Then Exit For
    q = p + diff: r = q + diff
    sq = p3 + q * q * q + r * r * r + s * s * s
    sr = Int(Sqr(sq) + 0.5)
    If sr * sr = sq Then
      Text1.Text = Text1.Text & p & Str(q) & Str(r) & Str(s) & crlf
      DoEvents
    End If
   End If
  Next diff
 Next p


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

Function gcd(a, b)
  x = a: y = b
  Do
   q = Int(x / y)
   z = x - q * y
   x = y: y = z
  Loop Until z = 0
  gcd = x
End Function


  Posted by Charlie on 2015-09-02 15:08:39
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