Is the number 100k+11 ever a square?
Solution 1:
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.
Solution 2:
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...