When is $ 999\cdots$ a perfect square? I'm interesting to look more about property of  this number $ 999\cdots$ , for even digits which form that number it's clear that's not a perfect square for example :$ 99=10^2-1,9999=10^4-1,\cdots$ , Now my question is: what about if the numbers of digits of  $ 999\cdots$ is odd ? how i proof or disproof that is a perfect square or nor ?
Note: The trivial case is for $n=1$  which is $9$ 
 A: The only number consisting entirely of nines that's a perfect square is $9$ itself.
Suppose $999$ were a perfect square.  It would have the form $(2n+1)^2$ with$n$ a whole number.  Therefore
$(2n+1)^2=4n^2+4n+1=999$
$4n(n+1)=998$
So $998$ has to be a multiple of $4$, but that doesn't work because the last two digits $98$ would have to be a multiple of $4$ and they aren't really so.  So, the second equation above has no whole number solutions, the first equation can't have any such solutions either and $999$ fails to be a perfect square.
The same thing happens with any number ending with $99$, since the last two digits are what ultimately decides what can be a multiple of $4$.  Once we have two nines at the end, therefore, there can never be a perfect square.
A: Any odd perfect square is of form $4k+1$ for some $k$. Now let $$a^2=10^{n}-1$$if $n\ge 2$ we have $$a^2=10^2\cdot10^{n-2}-1=4k-1$$which can't be a perfect square. So the only perfect square may be among those numbers with $$0\le n<2$$or $$0,9$$which are the only solutions.
A: Write: $$10^{2n+1}-1 = a^2$$
so $a=2k+1$ and now we have $$2^{2n+1}5^{2n+1}= 2\underbrace{(2k^2+ 2k+1)}_{\rm odd}$$
so $2^{2n+1}=2$ and thus $n=0$. Finally we have $9 = a^2$ so $a=3$.
A: To make $\underbrace{999\cdots 9}_{n\text{ many}}$ a perfect square, we need $9\cdot\underbrace{111\cdots 1}_{n\text{ many}} $ a perfect square. As, $9$ is already a perfect square, we need $\underbrace{111\cdots 1}_{n\text{ many}}$ to be a perfect square. But, observe that, $$11\equiv 3\pmod{4}\\ 111\equiv 3\pmod{4} \\ 1111\equiv 3\pmod{4}$$ in general, $$\underbrace{111\cdots 1}_{n\text{ many}}=2\underbrace{77\cdots 7}_{n-2\text{ many}}×4+3\equiv 3\pmod{4}$$
But, any odd perfect square can be of the form $4k+1$. Hence, none of the numbers is a perfect square.  
A: Write the number from the form
$a=\underset{n}{\overline{\underbrace{9999...9}}_{10}}=10^{n}-1 $ 
a perfect square$ \Rightarrow a=k^{2}\Rightarrow 10^{n}-1=k^{2}\
\Rightarrow 10^{n}=k^{2}+1$
Our $(k^{2}+1)$ is not a perfect square because it is sandwiched between the squares of two consecutive numbers   $ k^{2}$  and $ (k+1)^{2}$   Which $k^{2}\lt k^{2}+1\lt (k+1)^{2}$
$k^{2}\lt 10^{n}\lt (k+1)^{2}$
For $a$ to be a perfect square, $10^{n}$ must not be a perfect square and be in the form of $k^{2}+1$
We study two cases (even and odd)
$n=2\lambda,n=2\lambda+1:\lambda\in N$
1) $n=2\lambda\to 10^{n}=\left( 10^{\lambda} \right)^{2}$  This does not fulfill the condition $k^{2}\lt 10^{n}\lt (k+1)^{2}$(There is no perfect square between the )
2)$n=2\lambda+1\Rightarrow k^{2}\lt 10^{2\lambda+1}\lt (k+1)^{2}$
$3^{2}\lt 10\lt 4^{2}\to 3^{2}.10^{2\lambda}\lt 10^{2\lambda+1}\lt 4^{2}.10^{2\lambda}$
$\left( 3\times 10^{\lambda} \right)^{2}\lt10^{2\lambda+1}\lt \left( 4\times 10^{\lambda} \right)^{2}$
$k^{2}\lt 10^{2\lambda+1}\lt (k+1)^{2}$
$k=3\times 10^{\lambda},k+1=4\times 10^{\lambda}\Rightarrow 3\times 10^{\lambda}+1=4\times 10^{\lambda}$
$\to 10^{\lambda}=1\Rightarrow \lambda=0$
$(\lambda=0)\to a=10^{n}-1=10^{2\lambda+1}-1=10-1=9$
Abstract / The number $999\cdots$ It is impossible to be a perfect square if the number of its digits is an even number, and it can be a perfect square in the case of the number of its digits being odd in one case, which is the number is $9$
A: Any number that is a square, and does have more then one digit, consists of at least one even digit.
So a square can not have only odd digits.
With other words $999\dotso$ can not be a square.
