Question

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

Answer 1

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

Both `(10k+1)^2` and `(10k+9)^2` must have an even `10`'s digit, but that would not match `1`.

Explanation:

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`11, 111, 1111, ...`

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

Suppose `n^2 = 100m+11` for some integers `m` and `n`

Let `k` and `c` be integers such that `n = 10k+c` and `0<=c<10`

Then `n^2 = (10k+c)^2 = 100k^2+20kc+c^2 = 100m+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 = 100m + 11`