Find all integral ordered pairs $(n,k)$ such that $\left\lfloor\frac{n^2+18n+10}{2}\right\rfloor = k^2$. I had two problems that I want to solve. The first one was easy, but the second one... not so much:


First Problem:

Find all values of $n$ such that$$\frac{(n+1)(n+9)+8n+1}{2} = n^2.$$

My Attempt:
$$(n+1)(n+9) = n^2 + 10n + 9.$$ Therefore, the problem is asking me to find all values of $n$ such that $$\begin{align} \frac{n^2 + 18n + 10}{2} &= n^2 \\ \\ \Leftrightarrow n^2 + 18n + 10 &= 2n^2 \\ \\ \Leftrightarrow 18n + 10 &= n^2 \\ \\ \Leftrightarrow -n^2 + 18n + 10 &= 0.\end{align}$$ By the Fundamental Theorem of Algebra (FTOA), I know that there must exist only two distinct solutions for $n$. Then, using the quadratic formula, I obtain that
$$\begin{align} n &=\frac{-18\pm\sqrt{18^2 - (-4\cdot 10)}}{2} \\ \\ &= \frac{18}{2}\pm\frac{\sqrt{324 + 40}}{2} \\ \\ &= 9\pm\sqrt{364}.\end{align}$$ These are the only solutions, and since $364$ is not a square number, then these solutions are also irrational.



Second Problem:

Find all ordered pairs $(n,k)\in\mathbb{Z}^2$ such that $$\left\lfloor\frac{(n+1)(n+9)+8n+1}{2}\right\rfloor=k^2$$ for which $n\neq k$ and we denote by $\left\lfloor x\right\rfloor$ rounding $x$ to the lowest integer.

How must I approach this problem? Because we have the $\left\lfloor\ldots\right\rfloor$ function, I assume that in at least one case, $(n+1)(n+9)+8n+1$ is odd. Therefore, in at least one case, $n$ must be odd. By trial and error, it seems like there exist only three solutions: $(2, 5), (3, 6), (4, 7)$.
My Attempt:
Executed Trial and Error and expanded the numerator, but it did not help.

Thank you in advance.
Edit (Another Attempt):
If $n$ is even, then $\exists a \in\mathbb{Z}$, the set of all integers, such that $n = 2a$. Simplifying results in $2a^2 + 24a$ $+ \, 5 = k^2$. Since the quadratic is congruent to $1$ modulo $4$ (or $a$ is a quadratic residue modulo $4$), then it definitely can be a square.
When $k=0$ then $a=-6\pm\sqrt{536}\notin\mathbb{Z}$ so $k\neq 0$.
When $k = 1$ then $a=-6\pm\sqrt{544}\notin\mathbb{Z}$ so $k\neq 1$.
When $k=2$ then $a=-6\pm\sqrt{568}\notin\mathbb{Z}$ so $k\neq 2$.
Now I go back to trial and error....

The floor function seems annoying so lets get rid of it, letting $n=2m$ we get:
$\frac{(2m+1)(2m+9)+16m +1}{2}=k^2$.
So $4m^2+36m+10=2k^2$ or $(2m+9)^2 = 2k^2+71$

letting $n=2m+1$ we get
$\frac{(2m+2)(2m+10)+16m+9-1}{2}=k^2$
So $4m^2+40m+28 = 2k^2$ or $(2m+10)^2=2k^2+72$

Hence the problem reduces to finding all values of $a^2-2k^2=71$ or $a^2-2k^2=72$ which can be done with Pell equations.
