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

b) Find all strictly positive integers $n$ that are not multiples of 10 such that when writing $n^{2}$ and removing the last two digits, the resulting number is still the square of an integer (we consider a number with zero digits to be zero).
Solution to exercise 4 a) Any multiple of 10 works. Indeed, a strictly positive multiple of 10 can be written as $10 k, k \in \mathbb{N}^{*}$. Then its square is $100 k^{2}$, 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 $d$ be the number such that $d^{2}$ is equal to $n^{2}$ minus its last two digits. Then $n^{2}$ 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 $n^{2}$ 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 $\mathrm{n}^{2} \leqslant(10 \mathrm{~d})^{2}+99$, we have $10 \mathrm{~d} \leqslant 48$. Since $\mathrm{d}$ is an integer, $\mathrm{d} \leqslant 4$, thus $10 \mathrm{~d} \leqslant 40$ and $(10 \mathrm{~d})^{2} \leqslant 1600$. Therefore, $\mathrm{n}^{2} \leqslant 1699$, so $\mathrm{n} \leqslant 41$ 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 $d \leqslant 2$. If the last digit of $n$ is not 1, 2, 3, or 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.