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
how about 2019 instead of 2018? by Steven Lord)
assuming my initial analysis is correct (still needs proof) the determinant for n is
2017^(n-1)*(2017+n)
if n-1 is even then 2017+n is square
thus 2017+n=k^2 or n=k^2-2017
for n-1 to be even, n is odd which then means k is even
so for even k we have n=k^2-2017 as a solution
the smallest integer value for n this gives is when k=46 giving n=99
if n-1 is odd then 2017*(2017+n) is square
if 2017(2017+n) is square then 2017 must divide 2017+n
so n=2017k and since n-1 is odd, n must be even thus k must be even
2017(2017+2017k)
2017^2(k+1)
so then k+1 must be a square or k=t^2-1
so t must be odd to make k even
smallest solution here is when t=3 giving k=8 and n=2017*8>99
so from this analysis the smallest solution is when n=99
giving a determinant of
2017^98*(2017+99)=2017^98*2116=2017^98*46^2=(2017^49*46)^2
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Posted by Daniel
on 2018-06-15 21:05:50 |
re: how about 2019 instead of 2018?
