 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Looking for a square result (Posted on 2018-06-14) 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 No Rating Comments: ( Back to comment list | You must be logged in to post comments.) 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 Please log in:

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