Day 19: Counting Squares

We prove that $\underbrace{111\cdots 1}_{k}\underbrace{555\cdots 5}_{k-2}56$ is always a perfect square ( $k\geq 2$ ).

We notice that it can be written in the form: $\frac{10^{2k}-1}{9} + 4 \times \frac{10^{k}-1}{9} + 1 = \frac{10^{2k}-1 + 4 \times 10^{k}-4 + 9}{9} = \\ \frac{10^{2k} + 4 \times 10^{k} + 4}{9} = (\frac{10^k+2}{3})^2$ So it is a perfect square.

Also, $\frac{10^k+2}{3} \geq 34$ as it increases from $k=2$ .

Now all the squares in the set are $5$ more than these squares. But the next square numbers after these squares are always at least $34 + 35$ after them. Hence no numbers in the set are perfect squares, and the answer is $\boxed{0}$ as required.

As $10^n\equiv(-1)^n$ modulus 11, we have :

$\underbrace{111\cdots 1}_{2m}\underbrace{555\cdots 5}_{2m-2}61\equiv(-1+1-1\cdots +1)+(-5+5-5\cdots +5)-6+1\equiv 6$ modulus 11,

and

$\underbrace{111\cdots 1}_{2m+1}\underbrace{555\cdots 5}_{2m-1}61\equiv(-1+1-1\cdots -1)+(5-5\cdots +5)-6+1\equiv -1+5-6+1\equiv 10$ modulus 11.

But no square is congruent to 6 or 10 modulus 11: