B2. For how many integers $n$ with $1 \leqslant n \leqslant 800$ is $8 n+1$ a square?
B2. 39 If $k$ is an even number, then $k^{2}$ is even and thus never of the form $8 n+1$. If $k$ is odd, say $k=2 \ell+1$, then $k^{2}=(2 \ell+1)^{2}=4 \ell^{2}+4 \ell+1=4 \ell(\ell+1)+1$. One of the numbers $\ell$ or $\ell+1$ is even, and thus $4 \ell(\ell+1)$ is a multiple of 8, making $k^{2}$ of the form $8 n+1$.

We further see that $8 n+1$ is at least 9, the square of 3, and at most $8 \cdot 800+1=6401$. Therefore, we only need to determine how many odd squares lie between these values. We observe that $80^{2}=6400$ is still less than 6401, and that $81^{2}=6561$ is greater than 6401. The squares we are looking for are thus $3^{2}, 5^{2}, 7^{2}, \ldots, 79^{2}$. There are a total of 39 squares.