Repeated units!

Every integer can be written as $2n$ or $2n +1$ , thus every perfect square number can be written as $4n^2$ or $4n^2 + 4n + 1$ for some integer n.

$11111...$ can be written as $10^k + 10^{k-1} + ... + 10 + 1$ for positive integer k.

Proof by contradiction:

Let $a = \sqrt{11111...}$ , which then implies $a^2 = 11111... = 10^k + 10^{k-1} + ... + 10 + 1$ for positive integer k

Assume $a^2$ is a perfect square number, then $10^m + 10^{m-1} + ... + 100 + 10 + 1 = 4n^2 + 4n +1$ for some positive integers m and n. ( $a^2 = 4n^2$ is not regarded because $a^2$ is odd)

$10^m + 10^{m-1} + ... + 100 + 10 = 4n(n + 1)$ (subtract 1 on both sides)

$10^{m-1} + 10^{m-2} + ... 10 + 1 = \frac{4n(n+1)}{10} = \frac{2n(n+1)}{5}$ (divide both sides by 10)

$\frac{2n(n+1)}{5}$ has to be an odd integer, because $10^{m-1} + 10^{m-2} + ... 10 + 1$ is an odd integer, however, $\frac{Even}{Odd} = Even$

So $\frac{2n(n+1)}{5}$ is both even, and odd and that is a contradiction . Therefore no values of m and n can satisfy the equation.

Note that only $a^2 = 1$ is possible as dividing by 10 on both sides do not produce an integer (and the contradiction is avoided).

All numbers that end in $11$ are congruent to $3\text{ mod }4$ and thus cannot be perfect squares.

Let $R_n=\overbrace{11\cdots1}^{n\text{ }1\text{s}}$ . By observation, $R_2$ is not a perfect square. Note that, for $n\geq 3$ , $R_n = 100R_{n-2} + 11 \equiv 3 \bmod 4$ . However, $0^2\equiv 2^2 \equiv 0 \bmod 4$ and $1^2 \equiv 3^2 \equiv 1 \bmod 4$ . Thus no perfect square is congruent to $3\bmod 4$ and no integer consisting of repeated $1$ s (other than $1$ ) can be a perfect square.

If there was a square rep-unit, since it ends in 1, it has to be (10n +- 1)^2. But that would give us 100n^2 + 20n + 1. Since 2 is even, and an even multiplied to any number is still even, it is impossible for the tens to be 1.

$\sqrt{1}=1$ , now if any of the statements that follow that one are true, then there must be some number, which when squared, the rightmost two digits are 1 Since the last digit in the square number is 1, the rightmost digit can only be 1 or 9. The only digits that could possibly change the tens digit in the square number are the 2 rightmost digits. yet, no matter what digit you place in the tens digit, there will be an even number in the tens digit of the square number.

$11=4*2+3$ , $111=4*27+3$ ,...in a same way $111111111111111111...11=4*k+3$ for some positive integer. Therefore $11111111111111111111111111...11\equiv 3\bmod 4$ but we know that any perfect square integer is either congruent to $0$ or $1$ $mod 4$ .Hence 1 is the only integer in the above sequence.

A perfect square whose single digit is 1 cannot have an odd tens digit. Let 111...=a 2, a’s single digit must be 1 or 9. so a=10k+1 or a=10k+9. <1> a=10k+1, a 2=20k(5k+1)+1; <2> a=10k+9, a**2=20k(5k+9)+81; So a’s tens digit must be even.

For any perfect square ending with 1, the root integer must either be ending with 1 or 9. Except for integer 1 which is square of itself, the square of any number ending with 1 or 9 will generate an output ending with 01, 21, 41, 61 or 81 only. So, it is not possible for any other numbers consisting of all 1s to be a perfect square.

Suppose $x^2 = 111\ldots$ is such a solution. If its digits are $x = \overline{\cdots a_2a_1a_0}$ , we immediately see that $a_0$ must be 1, but then the second digit of $x^2$ is $2a_1$ , which cannot be 1.

Definitions:

$\overbrace{111...11}^{n \ no. \ of \ 1s}$ can be written as $\sum_{i=0}^{n}10^i$ .

Proof by contradiction:

Our assumption is that it's possible to write $k^2 = \sum_{i=0}^{n}10^i$ for some integer $k > 1$ and $n$ .

Our first lemma is that $k^2$ and by extension $k$ must always be odd , because $\sum_{i=0}^{n}10^i = 1 + \sum_{i=1}^{n-1}10^i$ , which is just $1 + 2m$ where $m$ is some integer.

Consider the squares of odd numbers: $1, 9, 25, 49,...$

Our second lemma is that squares of odd numbers can always be represented by the form $8a + 1$ for some integer $a$ . Short proof: We have $(2c + 1)^2$ for some integer $c$ and odd number $2c + 1$ , this equals $4c^2 + 4c + 1 = 4c(c+1) + 1$ , note that $c(c+1)$ is always even and can be written as $2c$ instead, therefore $(2c + 1)^2 = 4c(c+1) + 1 = 8c + 1$ .

Now for our assumption to be true, we need to be able to set $8a + 1 = \sum_{i=0}^{n}10^i$ for some integer $a$ and $n$ .

Remove $1$ from both sides, and divide by $8$ to get:

$a = \dfrac{-1 + \sum_{i=0}^{n}10^i}{8} =\frac{1 - 1}{8} + \frac{10}{8} + \frac{100}{8} + \frac{1000}{8} + ... =0 + 1.25 + 12.5 + 125 + ... = 0.75 + \overbrace{1 + 12 + 125 + ...}^{b}$

This is the same as $a = 0.75 + b$ where we have integer $a$ and $b$ , obviously this is a contradiction because we can't have an integer $a$ with a fraction, this means our assumption that there is some integer $k > 1$ and $n$ to write $k^2 = \sum_{i=0}^{n}10^i$ is wrong and therefore outside of the trivial case $k = 1$ there exists no such $k$ . $\square$

Every perfect square leaves remainder 0 or 1, when divided by 4, as it follows from @Johanan Paul

We can check last two digits of 111...11 for divisibility for 4. It leaves remainder 3, hence it can't be a perfect square!

Let any odd number be 2k-1. The square of this number is 4(k^2-k)+1 which is always 1 (mod 8).

Any number of the form 111...111 which is >100 is 111 (mod 8) a.k.a 7 (mod 8) and is therefore not a perfect square.

11 is known not to be a perfect square, and so we are done.

(we have incidentally proven that NO number ending 111 is a perfect square).

Bcuz any number is divisible by 1 which is the square root of 1