# Prove that none of $\{11, 111, 1111,\dots \}$ is the perfect square of an integer

prove that none of $\{11, 111, 1111 \ldots \}$ is the square of any $x\in\mathbb{Z}$ (that is, there is no $x\in\mathbb{Z}$ such that $x^2\in\{11, 111, 1111, \ldots\}$).

-
@DrammaFreak, not sure about your expectation. You may have a look into math.stackexchange.com/questions/418161/… – lab bhattacharjee Jun 13 '13 at 15:50

Hint: Perfect squares are not of the form $4k+3$, where $k$ is an integer.

Hint: For an even integer, $n=2j$, then $n^2 = (2j)^2 = ??$ For an odd integer, $n=2j+1$, then $n^2 = (2j+1)^2 = ??$.

-
this is intuitive but how do i prove this? – user45099 Feb 8 '13 at 19:34
do you know why Perfect squares are not of the form 4k+3 ? – Pechenka Feb 8 '13 at 19:35
@Pechenka Because $n^2 \bmod 4$ can only be $0$ or $1$ – Sasha Feb 8 '13 at 19:38
@DramaFreak If you explain what additional detail you need / don't understand, that might be better than simply posting a bounty. – Calvin Lin Jun 18 '13 at 23:02

Let suppose that there exists a number that squared gives $11 \cdots111$. Let $ba$ be its last two digits. Then either $a=1$ or $a=9$.

But if $a=1$, then the tens digit is $b + b \pmod{10}$ , which is even.

If $a=9$, , then the tens digit is $9 b + 8 + 9 b \pmod{10} = 18 b + 8 \pmod{10}$; which is also even.

Then, the tens digit cannot be $1$.

By the same reasoning, you can get the stronger result (see Philip Gibbs' answer) that a square cannot end with two odd digits.

-
Why did you use this approach? – Username Unknown Jun 12 '13 at 18:03
@DramaFreak What do you mean? Why not? – leonbloy Jun 12 '13 at 18:20
Where does $b+8+9b=1$ come from? $(xyzw...b9)^2$ has tens digit $(2\cdot9\cdot b+b^2)=8b+b^2\bmod 10$, but I don't see where that equates at all to the posed expression. The congruency to $8\pmod{10}$ suggests that every number ending in $9$ has a square that ends in $81$, which clearly isn't true. – Steven Stadnicki Jun 18 '13 at 22:32
@StevenStadnicki: Yes, there was an error (also in your formula), I hope it's better now. – leonbloy Jun 18 '13 at 23:52
@leonbloy Looks much better - I spaced that of course $b^2$ appears in the hundreds' place, not the tens'. – Steven Stadnicki Jun 19 '13 at 0:04

A square number can never end with two odd digits

If it did it would have to be the square of an odd number $x = 10a+b$ where $b$ is odd.

$x^2 = 100a^2 + 20ab + b^2$ so you just have to check for $x = 1,3,5,7$ or $9$ that the 10s digit is even and the rest follow.

-
downvoter explain? – Philip Gibbs Mar 12 '14 at 21:46

Every integer is of one of the forms $\color{green}4k$, $\color{green}4k+\color{red}1$, $\color{green}4k+\color{red}2$ or $\color{green}4k+\color{red}3$ with $k$ integer. The square of an integer has therefore one of the forms

• $(\color{green}4k)^2 = 16k^2 = \color{green}4(4k^2) = \color{green}4k_0$,
• $(\color{green}4k+\color{red}1)^2 = 16k^2+8k+1 = \color{green}4(4k^2+2k)+\color{red}1 = \color{green}4k_1+\color{red}{1}$,
• $(\color{green}4k+\color{red}2)^2 = 16k^2+16k+4 = \color{green}4(4k^2+4k+1)=\color{green}4k_2$ or
• $(\color{green}4k+\color{red}3)^2 = 16k^2+24k+9 = \color{green}4(4k^2+6k+2)+\color{red}1 = \color{green}4k_3+\color{red}1$.

That is, a perfect square is equivalent either to $\color{red}0$ or to $\color{red}1$ modulo $\color{green}4$.

On the other hand, $R_n=\underbrace{1\ldots1}_n=(10^n-1)/9$ for $n>1$ has the form

• $R_n=100R_{n-2}+11 = \color{green}4(25R_{n-2}+2)+\color{red}3 = \color{green}4k_r+\color{red}3$.

That is, for $n>1$, $R_n$ is equivalent to $\color{red}3$ modulo $\color{green}4$, which as shown above does not happen for perfect squares.

-

If $n\equiv11\pmod{100}$, then \begin{align} n &=100k+11\\ &=4(25k+2)+3\\ &\equiv3\pmod{4} \end{align} If $n$ is even, $n=2k$ and $n^2=4k^2$. Thus, $n^2\equiv0\pmod{4}$.
If $n$ is odd, $n=2k+1$ and $n^2=4(k^2+k)+1$. Thus, $n^2\equiv1\pmod{4}$

Therefore, whether $n$ is even or odd, $n^2\not\equiv3\pmod{4}$, and consequently, $$n^2\not\equiv11\pmod{100}$$

-

Well, consider an odd number $n=2k+1$. Then, $n^2 = (2k+1)^2= 4k^2+4k+1= 4(k^2+k)+1$.

So, $n^2 = 1 (mod \space 4)$, or, in other words, $n^2$ will always give you a remainder of 1 on division by 4, if n is odd.

On the other hand, if n is even, then, say, $n = 2k \Rightarrow n^2 = 4k^2 \Rightarrow n^2 = 0 (mod 4)$. Or, $n^2$ gives you a remainder of 0 on division by 4.

So, for any square, $n^2 = 0 (mod 4)$ if n is even, and $n^2 = 1 (mod 4)$ if n is odd.

Now, look at the numbers in the sequence. Firstly, note that 100 is divisible by 4. And, say, split each number into a multiple of 100 added to 11.

For example, $1111= 11000 + 11, 11111= 11100+ 11, \cdots$.

Now, each number in the sequence is equivalent to 11 (mod 4). But $11= 3 (mod 4)$, so, each number in the sequence is equivalent to 3, or gives a remainder of 3 on division by 4.

But, this is impossible for a square number, as we've shown above.

Therefore, no number in this sequence is a square.

-

Any odd square is congruent to 1 modulo 4 $(2n+1)^2 = 4n^2+4n+1 = 4(n^2+n)+1 \cong 1 \bmod{4}$

But 11 is congruent to 3 modulo 4.

Just as well any positive integer ending with 11 is not a perfect square.

-

If you examine the sequence formed by the last 2 digits of a square of an integer, or equivalently, look at the sequence $n^2 \pmod{100}$, it is easy to show that a square of an integer can only end with one of the following pairs of digits: 00, 01, 21, 41, 61, 81, 04, 24, 44, 64, 84, 06, 16, 36, 56, 76, 96, 09, 29, 49, 69, 89 or 25.

Each number in your sequence ends with $..11$, hence none of them can be squares.

From the above observation, it is also clear that many other sequences can also never contain a square, for example:

22, 222, 2222, ...

14, 414, 1414, 41414, ...

One could spend many happy hours creating such sequences.

-
I have been designed by the Sirius Cybernetics Corporation to take you, the visitor to the Hitch Hiker's Guide to the Galaxy, into these their offices. If you enjoy your ride, which will be swift and pleasurable, then you may care to experience some of the other elevators which have recently been installed in the offices of the Galactic tax department, Boobiloo Baby Foods and the Sirian State Mental Hospital, where many ex-Sirius Cybernetics Corporation executives will be delighted to welcome your visits, sympathy, and happy tales of the outside world.'' – Will Jagy Jun 18 '13 at 22:46
$$\begin{array}{l}\text{Little Jack Horner}\cr \text{Sits in a corner}\cr \text{Extracting cube roots to infinity,}\cr \text{An assignment for boys}\cr \text{That will minimize noise}\cr \text{And produce a more peaceful vicinity.}\end{array}$$ – Will Jagy Jun 18 '13 at 23:15

Let's first reformulate the claim: $$\nexists n \in \mathbb{Z}, k \in \{11, \space 111, \space \ldots \} : n^2 = k$$

Proof: First, consider all non-negative integers $n$ such that $n$ is even. That is, $$n = 2j \space \space \forall j \in \mathbb{Z}$$ For these $n$ we know that $n^2$ is of the form: $$n^2 = (2j)^2 = 2^2 \cdot j^2 = 4j^2$$ This means $n^2$ can be divided by 2 at least twice. Another way of saying this is that half of $n^2$ is even. However, $k$ is of the following form: $$k = 10a + 1 \space \space \forall a \in \{1, \space 11, \space \ldots \}$$ Since $10a$ is even, $10a + 1$ is not. However, since $n^2$ is divisible by 4, it is also even. Hence we know that: $$\nexists n \in \mathbb{Z}, j \in \mathbb{Z}, k \in \{11, \space 111, \space \ldots \} : n = 2j \space \land \space n^2 = k$$ That leaves us with all non-negative integers $n$ such that $n$ is uneven. Those can be written as: $$n = 2j + 1 \space \space \forall j \in \mathbb{Z}$$ In this case $n^2$ will have the form: $$n^2 = (2j + 1)^2 = 4j^2 + 4j + 1$$ We can see right away that this number is odd. But $k$ is odd too, so that is of no help. However, the other terms of $n^2$ all contain a factor of 4. That means $n^2 - 1$ is divisible by 4. Since that is even stronger than knowing a number is even, we should compare $n^2 - 1$ to $k - 1$ and see if $k - 1$ fits the same bill. Let's re-write this argument in Mathese and see if $k - 1$ is divisible by 4: $$n^2 = k \iff n^2 - 1 = k - 1$$ $$n^2 - 1 = 4j^2 + 4j = 4(j + 1) \implies 4 \space | \space n^2 - 1$$ Hence $n^2 = k \iff 4 \space | \space k - 1$. Let's try to reformulate $k$ in such a way that shows whether: $$k - 1 \space \text{is uneven or} \space k - 1 = 2j \space \text{for any uneven nonnegative integer} \space j \text{.}$$ If so, we have sadly made little progress. But if not, we have completed our proof.

$$k \in \{11, \space 111, \space \ldots \} \iff k - 1 \in \{ 11 - 1, \space 111 - 1, \space \ldots \}$$ $$\kern 47pt \iff k - 1 \in \{ 10, \space 110, \space \ldots \}$$ $$\kern 73pt \iff k - 1 \in \{ 10 \cdot 1, \space 10 \cdot 11, \space \ldots \}$$ Since $10$ is even but not divisible by 4, the other factor that composes $k - 1$ must be even for $k - 1$ to be divisible by 4 as well. We can write an equation that separates the factor $10$ from the factor that is an element of $\{1, \space 11, \space \ldots \}$. $$k - 1 = 10 \cdot j \space \forall j \in \{1, \space 11, \space \ldots \}$$ Since we know already that $\{11, \space 111, \space \ldots \}$ contains only uneven numbers and $1$ is uneven, we know the set $\{1\} \cup \{11, \space 111, \space \ldots \} \equiv \{1, \space 11, \space \ldots \}$ contains only uneven numbers as well.

Hence $k - 1 \nmid 4 \space \forall k \in \{11, \space 111, \space \ldots \}$.

Hence $\nexists n \in \mathbb{Z}, k \in \{11, \space 111, \space \ldots \} : n^2 = k$

Which is that what was to be proven.

-