Perfect square ending with $1$'s and $2$'s Is it true that for any $n$, there exists a perfect square whose last $n$ digits are only $1$ or $2$?
For example we have $1^2=1$, $11^2=121$, $511^2=261121$. But $111^2=12321$, so we cannot just use $1$'s.
 A: Your statement is equivalent to the statement that 
For $10^n$, that at least one of the numbers of the form $\overline{a_1a_2a_3 \dots a_n}$ where $a_i=1 \text{ or } 2$ for $1 \le i \le n$ is a quadratic residue.
Assume that this statement is true for $n$. 
Then we have to show that this statement is true for $n+1$. 
Assume $i^2 \equiv x \times 10^{n}+\overline{a_1a_2a_3\dots a_n} \pmod {10^{n+1}}$. Also note that since $2$ is not a quadratic residue $\pmod {10}$, $i$ is coprime to $10$. 
Then, note that $$(10^{n}k+i)^2 \equiv (2ki+x) \times 10^n + \overline{a_1a_2a_3\dots a_n} \pmod {10^{n+1}}$$
Thus, we have to show there exists such $k$ that $$2ki+x \equiv 1 \text{ or } 2 \pmod {10}$$ For fixed $x,i$, which follows from CRT. 
A: Justification of the fact that why you just cannot use 1 only

Rather no integer among 1,2,3,5,6,7,8,9 can be used. Only 4 is valid upto $n \le 3$ and  only $0$ is valid $\forall n \in \mathbb{N}$.

Note that 2,3,7,8 are never the last digit of perfect squares.
Let $m^2 = d \cdot 10^2 + 11 \equiv 3 (mod \ 4)$ which is not possible.
Same reasoning for 55, 66, 99
So only 44 and 00 survives.
Done!
Now we will see that $444$ and $000$ also survives as last digits. For the validity of $444$ we need to use Chinese Remainder Theorem. (The most trickiest of this whole proof thing). 
But $4444$ wont be possible as $m^2=d\cdot 10^4 + 4444 \equiv 12 \ (mod \ 16)$ which is not a valid as $x^2 \equiv 0,1,4,9 \ (mod 16)$  
