 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  One square implies many (Posted on 2018-03-28) 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 No Rating Comments: ( Back to comment list | You must be logged in to post comments.) 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 Please log in:

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