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Please check my proof. Thank you!

Proof: $11,111,1111,...$ can all be written as follows $\underbrace{111...}_{\text{k times}}=1+10(\sum_{i=0}^{k-2}10^n)$

Let us assume $1+10(\sum_{i=0}^{k-2}10^n)=s^2$ where $s\in\Bbb{Z}$.

Then this means $s^2|1$ and $s^2|10$. The only possible $s^2$ is then $1$.

It is obvious that $1$ does not work. So this means there is no $s$ such that $1+10(\sum_{i=0}^{k-2}10^n)=s^2$. So we conclude that none of $11,111,1111,...$ are squares of an integer.

Edit: Once again... this proof is wrong. Please look at the answers below.

Correct Attempt: I shall try induction. We see that $11\cong3(\text{mod 4})$ . Now assume that $\underbrace{111...}_{\text{k times}}\cong3(\text{mod 4})$

Then for $\underbrace{111...}_{\text{k+1 times}}$ we see that the last dividend in the long division is $31$. So the largest possible last digit is $7$ and $7\times4=28$ and $31-28=3$. The remainder is therefore $3$. And so, $\underbrace{111...}_{\text{k+1 times}}\cong3(\text{mod 4})$. However, we know that square numbers(mentioned to me by https://math.stackexchange.com/users/279515/brahadeesh ) are either $0$ or $1$ in $\text{mod 4}$. So we conclude that all of them cannot be perfect squares.

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    $\begingroup$ What do you mean in the sentence: "$s^2|1$ and $s^2|10$. The only possible $s^2$ is then $1$"? $\endgroup$ Jul 3, 2020 at 8:11
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    $\begingroup$ It's not true that, if $c \mid a + b$ then $c \mid a$ and $c \mid b$. For example, take $a = b = 1$ and $c = 2$. $\endgroup$
    – user804886
    Jul 3, 2020 at 8:13
  • $\begingroup$ $s^2$ divides 1 and $s^2$ divides 10. $\endgroup$
    – Khant Rain
    Jul 3, 2020 at 8:13
  • $\begingroup$ How about $10^2$? $10^2$ divides $1$ and $10^2$ divides $10$. $\endgroup$
    – Toby Mak
    Jul 3, 2020 at 8:14
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    $\begingroup$ The update looks kind of alright. Note that the method of induction is not really necessary here - specifically, how did you find out that "the last dividend in the long division is $31$" when considering the number with $k+1$ ones? You did not need to use the induction hypothesis anywhere to conclude that, I'm sure. So, you will have directly shown that all these numbers are congruent to $3$ modulo $4$ when you complete the justification that the remainder is $31$. $\endgroup$
    – user279515
    Jul 3, 2020 at 9:37

1 Answer 1

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Let us assume $1+10(\sum_{i=0}^{k-2}10^n)=s^2$ where $s\in\Bbb{Z}$.

Then this means $s^2|1$ and $s^2|10$.

I don't follow this implication, how do you argue that $s^2 \mid 1$ and $s^2 \mid 10$? Indeed, as @user804886 notes in a comment under your question, $c \mid a + b$ does not imply that $c \mid a$ and $c \mid b$.


One way to prove this is to note (exercise!) that a square is always congruent to either $0$ or $1$ modulo $4$. But the numbers in your sequence are all congruent to $3$ modulo $4$, so none of them can be a square.

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