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My problem is well described in title. I know that it can be proved as follow:

Since every number in the sequence is of the form $4x+3$ and perfect square does not exist in such form so none is a perfect square.

But I need to prove in a different way, a way different from modular arithmatic. Any ideas??

Martin Sleziak
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super saiyan
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    The first integer in the sequence is a perfect square, by the way... – Steve Kass Jan 08 '17 at 18:22
  • You should remove 1 from sequence – arberavdullahu Jan 08 '17 at 18:22
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    Using the last two digits is equivalent to doing the $\bmod 4$ thing, lets try to keep it real please. – Asinomás Jan 08 '17 at 18:40
  • Ok, how's this (for half the cases anyway): Your numbers can be written as $a_n=\frac 19\times \left(10^n-1\right)$. Thus $a_n$ is a square iff $10^n-1$ is. But if $n=2k$ is even then of course $10^n$ is a perfect square and there can't be two squares in a row (other than $0,1$). – lulu Jan 08 '17 at 18:51
  • that's cool, I was trying the same thing but I don't think it can be extended to all cases. – Asinomás Jan 08 '17 at 18:52
  • @JorgeFernándezHidalgo Yeah, so far I'm failing there as well. – lulu Jan 08 '17 at 18:53
  • @Watson: No it's not! Read the question (and the comment-thread that follows) a little more carefully. Why always rush to close??? – barak manos Jan 08 '17 at 19:06
  • @barakmanos : Does this answer use modular arithmetic? – Watson Jan 08 '17 at 19:09
  • @Watson: No, but the accepted answer does. And in any case, a question should be closed as duplicate if a duplicate question exists, not if a duplicate answer (possibly) does. – barak manos Jan 08 '17 at 19:10

3 Answers3

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first assume $1111111111...11$ is a perfect square :

it is odd number like $(2k+1)^2$ so $$1111111...1111=(2k+1)^2 \\ 111111...111=4k^2+4k+1\\111111...111-1=4k(k+1)=8q\\ 111111...110=8q \\q=\frac{111111...110}{8}=\frac{55555.555}{4} $$this is paradox .because $q$ must be a natural number . so, there is no perfect square in $11111...1111$ numbers

Khosrotash
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Assume that it is a perfect square. Then, since the last digit is $1$, the number is of the form $(10n+1)^2$ or $(10n+9)^2$. But then, in either case, the tens digit would be even---a contradiction.

John Bentin
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Proof by contradiction:

Suppose that one of the elements in the sequence is a perfect square.

Let $n$ denote the root of that element.

The unit digit of $n$ must be either $1$ or $9$.

Observe (or calculate it manually if you don't trust me) that:

  • $\not\exists{n}\in\{01,11,21,31,41,51,61,71,81,91\}\text{ such that the last $2$ digits of $n^2$ are $11$}$
  • $\not\exists{n}\in\{09,19,29,39,49,59,69,79,89,99\}\text{ such that the last $2$ digits of $n^2$ are $11$}$

Any other digits of $n$ surely have no impact on these last $2$ digits.

Therefore no element in the sequence is a perfect square.

barak manos
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    @supersaiyan What exactly does "different from modular arithmetic" mean? Do you consider this proof to be using arithmetic modulo $100$? – Erick Wong Jan 08 '17 at 18:36
  • @ErickWong: I am not supersaiyan. – barak manos Jan 08 '17 at 18:36
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    Of course you aren't, I am trying to ask OP to clarify the question using this as an example of where the question is murky. – Erick Wong Jan 08 '17 at 18:37
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    @ErickWong: Haha, OK, I got it now. I just realized that it was the OP's username. And I also realized that one could interpret my answer to be using modular arithmetic, so I understand where your question to the OP stems from. Thanks. I will wait for his or her response, and decide whether or not I should delete this answer (too). I think that this "no modular arithmetic" restriction is somewhat subjected to interpretation. – barak manos Jan 08 '17 at 18:40