Find all integral ordered pairs $(n,k)$ such that $\left\lfloor\frac{n^2+18n+10}{2}\right\rfloor = k^2$.

Question

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....

Answer 1

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.

Answer 2

A different approach to obtaining the Pell equations:

Note that $(n+1)(n+9)+8n+1=n^2+18n+10=(n+9)^2-71$ Let's write $N=n+9$. Then

$$\begin{align} \left\lfloor(n+1)(n+9)+8n+1\over2 \right\lfloor=k^2 &\iff\left\lfloor N^2-71\over2\right\rfloor=k^2\\ &\iff k^2\le{N^2-71\over2}\lt k^2+1\\ &\iff71\le N^2-2k^2\lt73\\ &\iff 71=N^2-2k^2\quad\text{or}\quad 72=N^2-2k^2 \end{align}$$

Note that $71=11^2-2\cdot5^2$ and $72=12^2-2\cdot6^2$, so you do get solutions for both.