# Prove that none of the integers $11,111,1111,...$ are squares of an integer.

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.

• What do you mean in the sentence: "$s^2|1$ and $s^2|10$. The only possible $s^2$ is then $1$"? Jul 3, 2020 at 8:11
• 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 $c = 2$. Jul 3, 2020 at 8:13
• $s^2$ divides 1 and $s^2$ divides 10. Jul 3, 2020 at 8:13
• How about $10^2$? $10^2$ divides $1$ and $10^2$ divides $10$. Jul 3, 2020 at 8:14
• 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 $31$.
– user279515
Jul 3, 2020 at 9:37

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.