No members of a sequence is a perfect square Prove that no member of the sequence 11, 111, 1111, ... is a perfect square.
I noticed that the first four terms of the sequence (above) are none of them are a perfect square. But I do not know what to do afterwards.
 A: HINT:
As $\underbrace{11\cdots11}_{n\text{ terms }}$ is odd, its square root must be odd
$$\text{Now, }(2a+1)^2=4a^2+4a+1\equiv1\pmod 4$$
$$\text{But, }\underbrace{11\cdots11}_{n\text{ terms }}\equiv11\pmod{100}\equiv-1\pmod 4\text{ for }n>1$$
A: Since this particular question has been discussed before, let me add five coins by proving that each number of the form $\overline{aaaa....aaaa}$ can not be a perfect square. Indeed, suppose that $\overline{aaaa....aaaa}=k^2,$ in other words
$$k^2=a\cdot\frac{10^n-1}{9}.$$
Rewriting the last inequality we end up with
$$a(10^n-1)=(3k)^2.$$
Considering both sides modulo $5,$ we get that $a=0,1,-1(\mod 5).$ Note, that $a\ne 5,$ because LHS is not divisible by $5.$ Considering both sides modulo $4,$ we have that $a=0,-1 (\mod 4)$ because $n\ge 2.$ Combining last two observations we end up with this only possibility $a=4.$ Then, $k=2k_1$ and we end up with $$10^n-1=(3k_1)^2$$
which is impossible modulo $4$ for $n\ge 2.$    
A: There is no proof it holds in general, because in base 3, the numbers $2^2=11$ and $102^2=11111$ are both square, and in base 7, the number $26^2=1111$ is square. 
In decimal, the numbers reduces to $3$ mod 4, which no square produces.
One might note also that Shank's prime $487$ is one of two known primes in decimal, that if $p$ divides a rep-unit, then so does $p^2$.  The other prime is $56598313$.  3 also divides its period (to complete the picture), but its period is 1, which means that $3^2|10^x-1$ for all x, but not $(10^x-1)/(10-1)$. 
