Second Round

We want $8n+1=k^2$ for some integer $k$.

Then $8n=k^2-1=(k-1)(k+1)$.

So $n=\frac{(k-1)(k+1)}{8}$.

We require $n$ to be an integer with $1\le n\le 800$.

Let us analyze when $n$ is an integer.

Note that $k$ must be odd, since $k^2\equiv 1(\mathrm{mod}8)$ only when $k$ is odd.

Let $k=2m+1$ for integer $m\ge 1$.

Then:

$k-1=2m$, $k+1=2m+2$

So $(k-1)(k+1)=2m\cdot (2m+2)=4m(m+1)$

Therefore,

$n=\frac{4m(m+1)}{8}=\frac{m(m+1)}{2}$

We require $1\le n\le 800$.

So $1\le \frac{m(m+1)}{2}\le 800$

Multiply both sides by $2$:

$2\le m(m+1)\le 1600$

Now, $m(m+1)$ increases rapidly. Let's find the largest $m$ such that $m(m+1)\le 1600$.

Solve $m^2+m-1600\le 0$

The positive root is $m=\frac{-1+\sqrt{1+6400}}{2}=\frac{-1+\mathrm{80.00625...}}{2}\approx 39.5$

So $m$ can be at most $39$.

Now, $m$ must be a positive integer such that $m(m+1)\ge 2$.

For $m=1$, $1\cdot 2=2$.

So $m$ runs from $1$ to $39$ inclusive.

Thus, the number of such $n$ is $39$.

Answer: $39$