Last Week of Classes

This is the last week of class for first semester and let me finish it off with an interesting question!

Source: http://www.math.osu.edu/~maharry.1/ProblemsOfTheMonth/ProblemsOfTheMonthJanuary2006.html

Find the smallest number (other than 1) in the sequence {1 , 11 , 111 , 1111 , 11111, ...} that is a perfect square (like 25 or 36 or 144) or prove that none of them are perfect squares.

11 is not a perfect square
111 is not a perfect square
1111 is not a perfect square
11111 is not a perfect square
111111 is not a perfect square

…

Intuition:

                I think none of them are perfect squares.

                The recursive function of the sequence (i^th term):

                f(i) =      1                              if i = 0
                                f(i-1) + 10ⁱ            if i > 0

For the (i + 1)th term, with 1 as the first term. Each of these is a finite geometric series, so we have:

Now we need to know what perfect squares are.

We have two cases, odd and even.

For odd integer:

Let n be an arbitrary odd integer, where n = 2m+1
(2m+1)²  = 4m²+4m+1
                   = 4(m²+m)+1
                   = 4k+1                #where k = m²+m

For even integer:

Let n be an arbitrary even integer, where n = 2m

(2m)² = 4m²

          = 4k, where k = m²

So perfect squares can be written in forms 4k+1 or 4k; which k is an arbitrary integer larger than or equal to zero.

Proof (By contradiction):

                Assume the sequence f(i) has a perfect square number. Then we can write f(i) in form of 4k + 1 or 4k. But the close form of f(i) = 

where k = 10ⁱ⁺¹

a contradiction! So the sequence f(i) contains no perfect squares.

 Conclude sequence f(i) contains no perfect squares by contradiction.

I am not sure if my explanation is correct, but I gave my best shot on this question without anyone’s help. Feel free to comment on my solution.

P.S. Happy studying for finals everyone~