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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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derivation of the formula | Comment 16 of 18 |
(In reply to Yes, that's it! by Steven Lord)

if d is the diagonal element and the matrix is Mn=


d 1 1...
1 d 1...
1 1 d...
...

if we subtract row 1 from all other rows and call c=d-1 we get
rid of all the ones in row 2 and onward:

 (c+1)  1    1...
    -c     c    0...
    -c     0    c...
...

adding columns 2 through n to column 1 gives:

(c+1+n-1)   1    1...
      0           c    0...
      0           0    c...
...

the determinant is the product of the diagonal elements alone
since all other permutations of elements contain a zero element. 

det(Mn)(d) = (c+n) c^(n-1) = (d-1+n) (d-1)^(n-1)  QED

(This was Daniel's formula which he used to prove det(M99)(2018) to be the first square)


Edited on June 26, 2018, 12:18 pm
  Posted by Steven Lord on 2018-06-16 14:02:01

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