Prove that none of $\{11, 111, 1111,\dots \}$ is the perfect square of an integer

Question

Please help me with solving this :
prove that none of $\{11, 111, 1111 \ldots \}$ is the square of any $x\in\mathbb{Z}$ (that is, there is no $x\in\mathbb{Z}$ such that $x^2\in\{11, 111, 1111, \ldots\}$).

Answer 1

Hint: Perfect squares are not of the form $4k+3$, where $k$ is an integer.

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Hint: For an even integer, $n=2j$, then $n^2 = (2j)^2 = ??$ For an odd integer, $n=2j+1$, then $n^2 = (2j+1)^2 = ??$.

Answer 2

Let suppose that there exists a number that squared gives $11 \cdots111$. Let $ba$ be its last two digits. Then either $a=1$ or $a=9$.

But if $a=1$, then the tens digit is $b + b \pmod{10}$ , which is even.

If $a=9$, , then the tens digit is $9 b + 8 + 9 b \pmod{10} = 18 b + 8 \pmod{10}$; which is also even.

Then, the tens digit cannot be $1$.

By the same reasoning, you can get the stronger result (see Philip Gibbs' answer) that a square cannot end with two odd digits.