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 1,3,8 ...extend! (Posted on 2017-01-26)
The set (1,3,8) has this peculiar feature:

Multiply any two of these numbers together and add 1 and you'll always get a perfect square:

1+ 1 × 3 = 4
1 + 1 × 8 = 9
1 + 3 × 8 = 25

a. Add to the above set a 4th integer, thus enabling 3 additional equalities.
b. Can you show why no 5th member maintaining the above feature exists?

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 computer finding | Comment 1 of 4
In order to work with set member 1, each such number except 1 has to be 1 less than a perfect square.

The program tests all such numbers up to 1 less than 10,000,000^2 for compatibility with 3 and 8:

DefDbl A-Z
Dim crlf\$

Form1.Visible = True

Text1.Text = ""
crlf = Chr\$(13) + Chr\$(10)

For rt = 4 To 10000000
DoEvents
n = rt * rt - 1
n1 = 3 * n + 1: n2 = 8 * n + 1
sr = Int(Sqr(n1) + 0.5)
If sr * sr = n1 Then
sr = Int(Sqr(n2) + 0.5)
If sr * sr = n2 Then
Text1.Text = Text1.Text & n & crlf
End If
End If
Next rt

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

End Sub

Only the number 120 works.

3*120 = 19^2 - 1
8*120 = 31^2 - 1

 Posted by Charlie on 2017-01-26 14:38:53

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