Exercise 4. a) Find a positive integer $n$ such that if you write $n^2$ and remove the last two digits, the resulting number is still the square of an integer.

b) Find all positive integers $n$ that are not multiples of 10 such that when you write $n^2$ and remove the last two digits, the resulting number is still the square of an integer (a number with zero digits is considered to be zero).
Solution to exercise 4 a) Any multiple of 10 works. Indeed, a strictly positive multiple of 10 can be written as $10 \mathrm{k}, \mathrm{k} \in \mathbb{N}^{*}$. Then its square is $100 \mathrm{k}^{2}$, and if we remove the last two digits of its representation, which are zeros, we get $\mathrm{k}^{2}$, which is indeed a perfect square.

b) If $n$ is strictly less than 10, its square has at most two digits, so by removing them, we get zero according to the convention in the statement. If $n$ satisfies the property without being a multiple of 10, $n^{2}$ is not a multiple of 100. Let $\mathrm{d}$ be the number such that $\mathrm{d}^{2}$ is equal to $\mathrm{n}^{2}$ with its last two digits removed. Then $\mathrm{n}^{2}$ and $(10 \mathrm{~d})^{2}$ are distinct perfect squares whose difference is at most 99, because the representation of $(10 \mathrm{~d})^{2}$ is obtained by replacing the last two digits of that of $\mathrm{n}^{2}$ with zeros. Since $(10 \mathrm{~d})^{2}<\mathrm{n}^{2}$, we have $10 \mathrm{~d}<\mathrm{n}$. But $10 \mathrm{~d}$ and $n$ are integers, so $n \geqslant 10 \mathrm{~d}+1$, thus $\mathrm{n}^{2} \geqslant(10 \mathrm{~d})^{2}+2 \times 10 \mathrm{~d}+1$. Since $n^{2} \leqslant(10 \mathrm{~d})^{2}+99$, we have $10 \mathrm{~d} \leqslant 48$. Since $\mathrm{d}$ is an integer, $\mathrm{d} \leqslant 4$, so $10 \mathrm{~d} \leqslant 40$ and $(10 \mathrm{~d})^{2} \leqslant 1600$. Thus, $\mathrm{n}^{2} \leqslant 1699$, or $\mathrm{n} \leqslant 41$ since $n$ is an integer. If the last digit of $n$ is not 1, $n \geqslant 10 d+2$ so $n^{2} \geqslant(10 d)^{2}+40 d+4$ and we find $\mathrm{d} \leqslant 2$. If the last digit of $\mathrm{n}$ is neither 1 nor 2, $n \geqslant 10 \mathrm{~d}+3$ and we find similarly $\mathrm{d} \leqslant 1$. If it is neither 1, nor 2, nor 3, nor 4, $n \geqslant 10 d+5$ and we find $d=0$. Thus, we still need to test $11,12,13,14,21,22,31$ and 41, which we quickly verify are solutions, in addition to $1,2,3,4,5,6,7,8$ and 9.