Doubt about whether repunits are square

Question

I have doubts in this exercise.

• Can a number $A= 111...11$, ($1000$ times the number $1$), be a square?
• Can a number $B= 111...11$, ($10431$ times the number $1$), be a square?
• Can a number $C= 111...11$, ($n$ times the number $1$), be a square?

I thought that if the number ends in $1$ it is odd, so I can write it in the form $A = 2k + 1$.

But in this way, I always find that it is not a square, regardless of the number of times that the number $1$ appears.

Is it okay to think the way I'm thinking?

Thanks.

Answer 1

As said in comments, a square cannot be $3\pmod{4}.$ (See the proof here.)

Also note that all the numbers in a), b), and c) end in $11$ and therefore are $3\pmod{4}.$ (Divisibility rule by 4 is to find if the last two digits are divisible by 4, if they are, then the entire number is.)

Therefore, any number of just $1$s other than $1$ cannot be a perfect square.

Answer 2

An odd number can be a square; for example $121=11^2$.

But if you try squaring even ($2k$) and odd ($2k+1$) numbers, you will see that squares must leave remainder $0$ or $1$ when divided by $4$.

The remainder when a number is divided by $4$ is the same as the remainder when the number comprising its last two digits is divided by $4$.

Therefore, the remainder when the repunit number $111\dots11$ is divided by $4$ is the same as the remainder when $11$ is divided by $4$, which is $3$, so $111\dots11$ cannot be a square.