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Looking for a square result (Posted on 2018-06-14) Difficulty: 3 of 5
For each positive integer n, let Mn be the square matrix (nxn) where each diagonal entry is 2018, and every other entry is 1.

Determine the smallest positive integer n (if any) for which the value
of det(Mn) is a perfect square.

No Solution Yet Submitted by Ady TZIDON    
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Solution computer solution | Comment 3 of 18 |
DefDbl A-Z
Dim crlf$, tri(100, 100)


Private Sub Form_Load()
 Form1.Visible = True
 Text1.Text = ""
 crlf = Chr(13) + Chr(10)
 
 maxsize = 10
 
 
 For sz = 1 To maxsize
   For i = 1 To sz
    For j = 1 To sz
      If i = j Or i = sz + 1 - j Then
        tri(i, j) = 2018
      Else
        tri(i, j) = 1
      End If
    Next
   Next
   Text1.Text = Text1.Text & mform(sz, "##0") & " " & det(sz, tri()) & Str(Sqr(det(sz, tri()))) & crlf
   DoEvents
 Next
 
 
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

Function det(sz, triangle())
  If sz = 2 Then
    xx = xx
  End If
  If sz = 1 Then
    det = triangle(1, 1)
    Exit Function
  End If
  ReDim matrix(sz - 1, sz - 1)
  tot = 0
  For col = 1 To sz
    newrow = 0
    For r = 2 To sz
      newrow = newrow + 1
      newcol = 0
      For c = 1 To sz
        If c <> col Then
          newcol = newcol + 1
          matrix(newrow, newcol) = triangle(r, c)
        End If
      Next
    Next r
    term = triangle(1, col) * det(sz - 1, matrix())
    If col Mod 2 = 0 Then term = -term
    tot = tot + term
  Next col
  det = tot
End Function


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

finds

 n det  square root
  1 2018 44.9221548904324
  2 0 0
  3 0 0
  4 0 0
  5 0 0
  6 0 0
  7 0 0
  8 0 0
  9 201964039878876 14211405.2745981
 10 0 0

I attribute the anomalous value at n=9 to rounding errors due to the large sizes that the numbers get to at such large matrices.

For n>1, the determinant seems always to be zero, which is a perfect square. So the sought smallest value of n is 2, where the determinant of the matrix is easily seen to be zero as the matrix consists solely of 2018's.

  Posted by Charlie on 2018-06-15 07:25:52
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