Estonian Mathematical Olympiad

The condition that the last $k$ digits of two numbers are the same is fulfilled if and only if the difference of these two numbers ends with exactly $k$ zeroes. Note that $n^2+1-2n=(n-1{)}^2$.

a. Let $n=100010001$, then the number $n-1$ ends with 4 zeroes and the fifth digit from the end is 1. Hence the number $(n-1{)}^2$ ends with 8 zeroes and the ninth digit from the end is 1. Hence this $n$ fits.

b. If a number ends with exactly $k$ zeroes, then the square of this number ends with exactly $2k$ zeroes. Hence $(n-1{)}^2$ cannot end with exactly 9 zeroes, since 9 is odd, and so there are no such integers $n$ that would fulfill the condition.