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I have to prove that the sequence $\{11,111,1111, \dots \} $ doesn't contain any perfect square numbers. I can realize it but I am unable to prove it. Please help.

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marked as duplicate by Hakim, Tom Oldfield, Grigory M, Davide Giraudo, Martin Sleziak May 29 at 19:07

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3 Answers 3

up vote 23 down vote accepted

$$\underbrace{11\cdots11}_{n \text{ digits}}\equiv11\pmod{100}\text{ for }n\ge2$$

$$\implies\underbrace{11\cdots11}_{n \text{ digits}}\equiv11\pmod4\equiv-1\text{ for }n\ge2$$

But for any integer $\displaystyle a,a\equiv0,\pm1,2\pmod4$

$\displaystyle\implies a^2\equiv0,1\pmod4$

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The first term of the sequence, $x_{0}=11$, is not a perfect square. The rest of the terms are constructed as follows: $x_{n+1}=10x_{n}+1$. Suppose $10x_{n}+1=a^2$ for some integer $a$. Then $10x_{n}=a^2-1=(a+1)(a-1)$. Since all the numbers are odd, $a$ must also be odd, so both $a+1$ and $a-1$ are even, which means that $10x_{n}$ is a multiple of $4$. However, $x_{n}$ is odd and $10$ is not a multiple of $4$, so this can't be true.

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Here's an arithmetic proof.

All of the numbers in your sequence are of the form $100j+11$ for integer $j$.

Imagine a decimal representation of all M-digit integers as $\sum_{m=0}^{M-1} 10^m k_m$ with $k_m \in \{0 .. 9\}$. The only integers whose squares end in 1 have $k_0 \in \{1, 9\}$. To have those integers' squares' tens place equal to 1, then by long-form multiplication, we would need to have: $1 \equiv 2 k_1 \pmod{10}$ (when $k_0 = 1$) or $1 \equiv 8 + 18 k_1 \pmod{10}$ (when $k_0 = 9$), which is impossible for any $k_1$, because in both cases, the tens place value must be even. No perfect squares can end in 11, 31, 51, 71, 91.

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What to do when $j=10$? –  Sagnik Saha May 29 at 6:16

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