# How do you show that the sequence 11, 111, 1111, 11111,... contains no perfect square number ?

Jul 16, 2015

If any of these numbers is square then it is the square of $10 k + 1$ or $10 k + 9$ for some integer $k$.

Both ${\left(10 k + 1\right)}^{2}$ and ${\left(10 k + 9\right)}^{2}$ must have an even $10$'s digit, but that would not match $1$.

#### Explanation:

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$11 , 111 , 1111 , \ldots$

Note that all of these numbers are of the form $100 m + 11$ for some integer $m$.

Suppose ${n}^{2} = 100 m + 11$ for some integers $m$ and $n$

Let $k$ and $c$ be integers such that $n = 10 k + c$ and $0 \le c < 10$

Then ${n}^{2} = {\left(10 k + c\right)}^{2} = 100 {k}^{2} + 20 k c + {c}^{2} = 100 m + 11$

Look at this modulo $10$ and modulo $20$.

${c}^{2} = 1$ modulo $10$. So $c = 1$ or $c = 9$.

We require ${c}^{2} = 11$ modulo $20$

But ${1}^{2} = 1$ and ${9}^{2} = 81 = 1$ modulo $20$

So neither value of $c$ works.

So our initial supposition is incorrect and there is no $m$ and $n$ such that ${n}^{2} = 100 m + 11$