Problem 2. For each non-zero natural number $n$, we denote by $p(n)$ the largest perfect square at most equal to $n$.

a) Determine the number of pairs of non-zero natural numbers $(m, n)$, with $m \leq n$, for which

$$
p(2 m+1) \cdot p(2 n+1)=400
$$

b) Determine the set $\left\{n \in \mathbb{N}^{*} \mid n \leq 100\right.$ and $\left.\frac{p(n+1)}{p(n)} \notin \mathbb{N}\right\}$.
Solution. a) There are three possible cases:

Case 1: $p(2 m-1)=1, p(2 n-1)=400$; we obtain $1 \leq 2 m-1<4$ and $400 \leq 2 n-1<441$, hence $m \in\{1,2\}$ and $n \in\{200,201, \ldots, 219\} \ldots . .1 \mathbf{p}$

Case 2: $p(2 m-1)=4, p(2 n-1)=100$; it follows that $4 \leq 2 m-1<9$ and $100 \leq 2 n-1<121$, hence $m \in\{3,4\}$ and $n \in\{51,52, \ldots, 60\} \ldots \ldots . .1 \mathbf{p}$

Case 3: $p(2 m-1)=16, p(2 n-1)=25$; then $16 \leq 2 m-1<25$ and $25 \leq 2 n-1<36$, hence $m \in\{9,10,11,12\}$ and $n \in\{13,14, \ldots, 18\} \ldots 1$ p

In the first case, 40 pairs are formed, in the second case, 20 pairs are formed, and in the third case, 24 pairs are formed; in total, there are 84 pairs as stated in the problem . ...................................................................1p

b) Let $n \in \mathbb{N}$. Denoting $p(n)=k^{2}$, we get $k^{2} \leq n \leq(k+1)^{2}-1$, hence $k^{2}<n+1 \leq(k+1)^{2}$. Therefore, $p(n+1) \in\{k, k+1\} \ldots \ldots \ldots \ldots \ldots$ 1p

Then $\frac{p(n+1)}{p(n)} \notin \mathbb{N}$ if and only if $p(n)=k^{2} \neq 1$ and $p(n+1)=$ $(k+1)^{2}$, i.e., $n=(k+1)^{2}-1, k \in \mathbb{N}, k \geq 2$. Since $n \leq 100$, it follows that $k \leq 9$.

The required set is $\{8,15,24,35,48,63,80,99\}$ $2 p$