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Please help me with solving this :
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 $x^2\in\{11, 111, 1111, \ldots\}$).

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@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

8 Answers 8

up vote 36 down vote accepted

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 $n^2 = (2j+1)^2 = ??$.

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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.

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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.

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

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, $$ n^2\not\equiv11\pmod{100} $$

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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.

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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.

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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.

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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.

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