B3. For how many of the integers $n$ from 1 to 100 is $8 n+1$ the square of an integer?
B3. 13

If $8n+1$ is the square of an integer, then that integer must be odd. Conversely, the square of an odd integer is always an eightfold plus 1. Indeed, if $k$ is odd, we can write $k=2\ell+1$ for an integer $\ell$. We see that

$$
k^{2}=(2\ell+1)^{2}=4\ell^{2}+4\ell+1=4\ell(\ell+1)+1
$$

Since either $\ell$ or $\ell+1$ is even, we see that $4\ell(\ell+1)$ is divisible by 8, so $k^{2}$ is an eightfold plus 1.

We only need to determine how many odd squares $x$ there are with $8 \cdot 1+1 \leqslant x \leqslant 8 \cdot 100+1$. These are the squares $3^{2}=9, 5^{2}=25$ up to $27^{2}=729$, since $29^{2}=841$ is greater than 801. The number of squares of the desired form is thus 13.