Let $x=\underbrace{11 \ldots 1}_{n}$, then $9x+1 =10^n$ and our number is: \[4x\cdot 10^n+8x-10^n -3x+1= (4x-1)(9x+1)+5x+1 = 36x^2\]

$N=\frac{10^n-1}{9} \cdot 4 \cdot 10^n + \frac{10^n-1}{9} \cdot 8 - (10^n + \frac{10^{n-1}-1}{9} \cdot 3 \cdot 10 +2) = \frac{4 \cdot 10^n - 8 \cdot 10^n + 4}{9} = (\frac{2}{3}(10^n-1))^2$.

pablock wrote: $N=\frac{10^n-1}{9} \cdot 4 \cdot 10^n + \frac{10^n-1}{9} \cdot 8 - (10^n + \frac{10^{n-1}-1}{9} \cdot 3 \cdot 10 +2) = \frac{4 \cdot 10^n - 8 \cdot 10^n + 4}{9} = (\frac{2}{3}(10^n-1))^2$. There is no difference between your proof and "number1's"

we can determine the numer of the rest, in the units we can see that it's always 8-2, so the unit of the result is always 6, then we can see that we will always have 88...8-33...3 where the number of 8s and 3s always are n-1, and then we will have that the second half of the final number is 55...56, also we will always have 4-1 in the digit numer n, so our final number should be 44...4355...56, with the number of 4s and 5s equal to n-1, if we calculate the roots of all posible number we will always have as result 66...6 with the amount of 6 equal to n and with the exception of n=1 because we will have the number 36 witch root is 6, can someone help me to justify the last part? ps: i'm sorry if i'd made gramaticall errors, english isn't my native languages