# 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
5 years, 7 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"

- 5 years, 7 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.

- 5 years, 7 months ago

OK Allright :)

- 5 years, 7 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.

- 5 years, 7 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.

- 5 years, 7 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

- 4 years, 9 months ago

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

- 4 years, 9 months ago

Yup, It can't be.

- 4 years, 9 months ago

absolutely.

- 4 years, 9 months ago