Is the number 100k+11 ever a square? Is the number 100k+11 ever a square?
Obviously all such numbers end in 11. Can a square end with an 11?
 A: Let $a^2=100\cdot k+11$. Then easily $a$ must be odd.
So $(2n+1)^2=100\cdot k+11$. It follows $2(n^2+n)=50\cdot k+5$, which is impossible.
A: Let's prove the somewhat more general result that $100k +\bar{ab}$ where $a$ and $b$ are odd digits cannot be a square. Our argument will be based on reductio ad absurdum:
$c^2=100k +10a+b$
$b$ is odd $\Rightarrow c \text{ is odd}$
$$\Rightarrow (2n+1)^2 = 100k+10(2l+1)+(2m+1)\\
\Rightarrow 2(n^2+n)=50k+5(2l+1)+m$$
The left hand side is even, therefor the right hand side should be even as well; this means that m should be odd.
$$\Rightarrow m=2j+1 \\
b<10\Rightarrow j=0\text{ OR }1\\
\Rightarrow n(n+1)=25k+5l+3+j$$
The left hand side can be $0,1,2$ modulo $5$; but the right hand side would be $3 \text { OR }4$ modulo $5$; which is a contradiction. 

To make the last statement more clear; we elaborate as:
$$n(n+1)\mod5=\begin{cases}
0 & n\equiv0\text{ OR 4}\\
1 & n\equiv2\\
2 & n\equiv1\text{ OR 3}
\end{cases}$$
A: Hint: Write $x=100y+z$; then $x^2=100(100y^2+2yz)+z^2$. Can you have $x^2=100k+11$? That is, $z^2\equiv 11\pmod{100}$.
A: A simple solution
$$(10x+y)^2 = 100x^2 + 20 xy + y^2$$
where $x$ is any number, and $0 \le y \le 9$


*

* The first term on right hand side must end in two zeros.

* The second term is a multiple of 20 and therefore must end in 00, 20, 40, 60, or 80

* For the square to end in 1 $y$ must be either 1 or 9

* In which case $y^2$ is either 01 or 81

* Thus 4. and 2. imply that the square must end in either 01, 21, 41, 61, or 81
 
A: All numbers of this form are congruent $11 \pmod 4 \equiv 3 \pmod 4 $. Now search one example of the very small list of residueclasses of that form which is also a square...
A: No. Only odd numbers ending with 01, 09, 21, 25, 29, 41, 49, 61, 69, 81, and 89 are squares. I will leave the proof to you.
A: Ends by 11 means, $n² \mod 100 = 11$ which also implies end by one $n² \mod 10 = 1$ 
this is true if and only if $(n \mod 100)^2 \mod 100 = 11$ and $(n \mod 10)² \mod 10 = 1$. 
Ergo you juste have to check squares of 1, 11, 21, 31, 41, 51, 61, 71, 81 and 91 and see none ends with 11.
