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.
(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
derivation of the formula
