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Arithmetic prime (Posted on 2019-04-25) Difficulty: 3 of 5
What is the smallest positive integer k such that k+1, 2k+1, 3k+1, 4k+1 are all primes?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution computer solution | Comment 1 of 3
DefDbl A-Z
Dim crlf$
Function mform$(x, t$)
  a$ = Format$(x, t$)
  If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
  mform$ = a$
End Function

Private Sub Form_Load()
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
 
 
For k = 2 To 20000
  good = 1
  For t = k + 1 To 4 * k + 1 Step k
    If prmdiv(t) < t Then good = 0: Exit For
  Next
  If good Then Text1.Text = Text1.Text & k & crlf
  DoEvents
Next

Text1.Text = Text1.Text & "  done"
 
End Sub
 
Function prmdiv(num)
 Dim n, dv, q
 If num = 1 Then prmdiv = 1: Exit Function
 n = Abs(num): If n > 0 Then limit = Sqr(n) Else limit = 0
 If limit <> Int(limit) Then limit = Int(limit + 1)
 dv = 2: GoSub DivideIt
 dv = 3: GoSub DivideIt
 dv = 5: GoSub DivideIt
 dv = 7
 Do Until dv > limit
   GoSub DivideIt: dv = dv + 4 '11
   GoSub DivideIt: dv = dv + 2 '13
   GoSub DivideIt: dv = dv + 4 '17
   GoSub DivideIt: dv = dv + 2 '19
   GoSub DivideIt: dv = dv + 4 '23
   GoSub DivideIt: dv = dv + 6 '29
   GoSub DivideIt: dv = dv + 2 '31
   GoSub DivideIt: dv = dv + 6 '37
 Loop
 If n > 1 Then prmdiv = n
 Exit Function

DivideIt:
 Do
  q = Int(n / dv)
  If q * dv = n And n > 0 Then
    prmdiv = dv: Exit Function
   Else
    Exit Do
  End If
 Loop

 Return
End Function

Finds

330
1530
3060
4260
4950
6840
10830
15390
18120

all work as values of k,

So the smallest such k is 330.

A slight modification goes out to 5k+1, etc.:

to 5k+1     10830
to 6k+1     25410
to 7k+1    512820 
to 8k+1    512820

That is, the smallest to include 7k+1 also goes as far as 8k+1.


  Posted by Charlie on 2019-04-25 14:50:25
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