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

1 Answer
Jul 16, 2015

Answer:

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#