Let the number(N, say) consists of digits D_{0},D_{1},D_{2},...,D_{n}where each D_{i}is from {0,6} and D_{0}must be 6 to be a (n+1) digited number. Now observe that both 0 and 6 are evenly divisible by 2.

Consider 2 cases:

1. D_{n }= 6

2. D_{n }= 0

In case 1 if the last digit of N is 6 i.e. D_{n}=6 then dividing it by 2 gives us a number (N/2) with last digit 3 and hence cannot be a perfect square since any further division of (N/2) will not yield any 2 from it i.e. N consists of a single 2 and hence it is a non-square number.

In case 2 if the last digit of N is 0 i.e. D_{n}=0 then let the least consecutive sub-sequence of 0s from D_{n}backwards be upto D_{i}where 1 <= i <= n,since D_{0}must be 6 to be a (n+1) digited number.

Therefore,from D_{0}to D_{i-1}the number consists ofmixesof 0s and 6s and from D_{i}to D_{n}the number consists of all 0s. Pictorially the view is as follows:

Now separate D_{0}to D_{i-1}as a string of 0s and 6s and D_{i}to D_{n}as a string of only 0s.

In other words the number is as follows: N = D_{0}...D_{i-1}X 10^{(n-i+1)}= 6...6 X 10^{(n-i+1)}

Now by previous argument we know that D_{0}...D_{i-1}can't be a perfect square and it also consists of a factor 6=2x3. So if the other part i.e "power of 10 is a perfect square[i.e if (n-i+1) is even]" then overall the number N can't be a perfect square.

If (n-i+1) is odd then 10^{(n-i+1)}can be factored as 2^{n-i+1}X 5^{n-i+1}where we see that the part 5^{n-i+1}still can't be a perfect square( as n-i+1 is odd and no other factors of 5 are there in the whole number ).

So in that case also this required result is proved.