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

1 Answer
Jul 16, 2015

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