How to prove that $11...111$ is not the sum of two perfect squares

Question

I'm stuck with this problem:

Show that $a=11...111$ is not the sum of two perfect squares. That is to say, there are no pair of integers ($b$ , $c$) so that $b^2+c^2=a$. I think I am supposed to use equivalence classes in some way, but I do not know how to approach it.

Answer 1

Since $a$ is odd, $b$ and $c$ have different parity: $b$ is even and $c$ is odd, say.

Then $b^2+c^2\equiv 1\pmod 4$, but $a\equiv 3\pmod 4$.

In fact, no sum of two squares ends with $11$.

Answer 2

It's easy to prove that every perfect square numbers, when is divided by 4, the remainder must be 0 or 1. And now the solution is clear.

Answer 3

$a$ is also not the sum of three perfect squares once it ends in at least three digits (all $1$). As $1000$ is divisible by $8,$ we then have $$ a = bcdefgh111 = 1000 w + 111 \equiv 7 \pmod 8 $$

Answer 4

$$a = \underbrace{1111\dots11}_{n} = \sum^{n-1}_{k=0}10^k= \frac{10^{n}-1}{9} $$

$$\frac{10^{n}-2}{9} +\frac{1}{9} = $$

$$\underbrace{\left(\frac{\sqrt{10^{n} - 2}}{3}\right)^2}_{b^2}+\underbrace{\left(\frac{1}{3}\right)^2}_{c^2}= a$$ $$b, c \notin \mathbb{Z}$$

Also $a$ is not the sum of two perfect squares. $\Box$