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1,3,8 ...extend! (Posted on 2017-01-26) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts 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$


Private Sub Form_Load()
 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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