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# OPC 2 Problem 2

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

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

Note by Rajdeep Dhingra
2 years, 3 months ago

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Don't get me wrong, but I feel like most of the problems are from "an excursion in mathematics" · 2 years, 3 months ago

Well I have heard a lot about that book. But these problems I have thought up and one is a famous one from IMO. · 2 years, 3 months ago

OK Allright :) · 2 years, 3 months ago

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 · 1 year, 6 months ago

You mean any number with last 2 digits 11, 31, 51, 71, 91 cannot be a perfect square? · 1 year, 6 months ago

absolutely. · 1 year, 6 months ago

Yup, It can't be. · 1 year, 6 months ago

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. · 2 years, 3 months ago

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. · 2 years, 3 months ago