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One square implies many (Posted on 2018-03-28) Difficulty: 2 of 5
Prove the following:
If there is one perfect square in an arithmetic progression, then there are infinitely many.

No Solution Yet Submitted by Ady TZIDON    
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BTW... | Comment 4 of 6 |
(In reply to solution assuming all members of the series are integers by Charlie)

The successive squares used to prove the theorem are not necessarily or even usually all the squares in the sequence. In the example given, m^2=25 leads to n^2=215296, but a computer search finds many intervening squares:


 square   k    square root

  144      7    12
  484     27   22
  841     48   29     where the square = 25+17*k
 1521     88   39    
 2116    123    46
 3136    183    56
 3969    232    63
 5329    312    73
 6400    375    80
 8100    475    90
... limited to first 10

  sq = 25 : diff = 17
  Do
    sq = sq + diff
    sr = Int(Sqr(sq) + 0.5)
    k = k + 1
    If sr * sr = sq Then
      Text1.Text = Text1.Text & Str(sq) & Str(k) & "    " & sr & crlf
      ct = ct + 1
    End If
    DoEvents
  Loop Until ct = 10




  Posted by Charlie on 2018-03-28 10:25:33
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