Well, 100 = 10^2, 10000 = 100^2,.....
S'that what you mean?
Could you clarify this? The way I am reading it, it is not correct. For example, 16 is a perfect square, yet it has a 1 in the tens place.
One way to look at the problem is to consider all of the squares modulo 100. It's a little computationally intensive, but it will do the job. I computed the list in Mathematica:
{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0}
We can see that 11 does not appear on the list. Therefore, any number of the form 11...1 (other than 1) cannot be a perfect square.
What I meant is this.
For any perfect square to end in 1, say x = y^2. 'y' can only by of the form 10k + 1 or 10k - 1. This is easy to verify.
For either of these cases - the tenth place in 'x' will be even. This can be proved easily.
So, my claim is any perfect square ending in 1 will have an even number in it's tens place. And, number of the form 1111....111 violate that and hence can't be a perfect square.
Aman_cc I understand that I need to look at the tenths place...but would I do something like this:
9^2 = 81
29 ^2 = 841
11 ^2 = 121
21^2 = 441
I thought the question was asking if all numbers such as 1,11,111,1111,11111 be a square of an integer.
Could you possibly show me some examples of your thinking because I think you might be right.
I don't think you got my argument right.
Step 1: Prove that any perfect square with digit 1 at it's one's place will have an even digit at it's ten's place.
Can you do this using the hints I provided?
Step 2: A number of the form 11...11111 is clearly a voilation of the argument above, hence, can't be a perfect square.
This completes the proof. But I think the method by Also sprach Zarathustra is pretty neat/better