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
re: computer solution by Charlie)
Right - I assumed only the main diagonal was 2018s - I wonder which way is right.
I also encountered roundoff errors. Going to higher precision I get
possibly that n=6, 8, 10 may be squares. But I think the polynomial form should be able to be expressed as a square, no?
program dee
implicit none
real*8 xn,xi,xj,det,d,rdet
integer n
d=2018.
print*,' n det sqrt(det)'
print*,' ----------------------------------------'
do n=3,10
xn=n
if(2*(n/2).eq.n)then
xi=xn/2.
det= d**xn - xi*d +(xi-1)
else
det= d**xn - xn*d + (xn-1)
endif
rdet=sqrt(det)
print*,n,det,rdet
enddo
end
n det sqrt(det)
----------------------------------------
3 8217943780.0000000 90652.875188821228
4 16583822756941.000 4072323.9995045825
5 33466154331639484. 182937569.49199769
6 6.7534699441268818E+019 8217949831.9999990
7 1.3628502347248049E+023 369168015234.90692
8 2.7502317736746564E+026 16583822760976.000
9 5.5499677192754564E+029 744981054744042.25
10 1.1199834857497871E+033 33466154331649568.
Edited on June 15, 2018, 9:37 am
re(2): computer solution
