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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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re: how about 2019 instead of 2018? | Comment 14 of 18 |
(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

  Posted by Daniel on 2018-06-15 21:05:50
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