\( 2. \text{Prove that no number in this sequence } 11,111,1111,11111,........... \\ \text{ is a perfect square} \)

Give different methods of solving this. This was a problem to the OPC 2. The OPC 2 has already ended.

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## Comments

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TopNewestDon't get me wrong, but I feel like most of the problems are from "an excursion in mathematics"

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Well I have heard a lot about that book. But these problems I have thought up and one is a famous one from IMO.

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OK Allright :)

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any no will have last two digits 11 and therefore cannot be a perfect square .i hope u get it .if u dont just ask

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You mean any number with last 2 digits 11, 31, 51, 71, 91 cannot be a perfect square?

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absolutely.

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Yup, It can't be.

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The numbers in the sequence can be written as \(111....108+3\) now since the last two digits, i.e. \(08\), is divisible by \(4\), the numbers are of the form of \(4k+3\). Now note that a perfect square, say \(a^2\) when written in modulo \(4\)\(\implies a^2=0,1,2\pmod4\) but not \(=3\pmod4\) which contradicts to the form of the numbers in the sequence.

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write

\( \underbrace {1 \cdots 1} _ {n \textrm{ times}}= a^2\)

Of course \(a\) must be odd (and greater than \(1\) since our sequence begins with \(11\), in other words \(n \geq 2\) ). Consider the even \(b = a -1\) . We have

\( \underbrace {1 \cdots 1} _ {n \textrm{ times}}= b^2 -2b +1\)

\( \underbrace {1 \cdots 1} _ {n-1 \textrm{ times}}\times 10= b^2 - 2b\)

\(b\) is even so \(4\) divides the LHS, but clearly not th RHS. Contradiction.

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