Question
Let $m$ and $k$ be positive integers. Let $f(m, km)$ be the number of rectangles (including squares) on a $m$ by $km$ checkerboard. Determine all ordered pairs $(m, k)$ for which $f(m, km)$ is a perfect square.
Progress
Because a $x \times y$ checkerboard has $\binom{x+1}{2} \times \binom{y+1}{2}$ rectangles, $f(m, km)=\binom{m+1}{2} \times \binom{km+1}{2}=\frac{km^2(m+1)(km+1)}{4}$.
Since $f(m, km)$ is a perfect square, $(km+k)(km+1)$ needs to be a perfect square, which means that $(km+k)(km+1)=x^2$ for which $x\in \mathcal{N}_*$. It's easy to find that $(m, 1)$ is a "family" of solution for this question.
I checked the sequence https://oeis.org/A082649 with somebody and found out that it had a relationship with this question. For example, the first smallest $k$ equals to 1, the second smallest $k$ equals to $4m+4$, the third smallest $k$ equals to $16m^2+24m+9$, and the fourth smallest $k$ equals to $64m^3+128m^2+80m+16$. I am not sure how this sequence is related to this question, but I believe that it may not be a coincidence.
Conjecture
From the sequence, one of my classmates formed a conjecture that may be proved.
Conjecture: For any positive integer $m$, let $$k=\frac{\sinh^2(n \sinh^{-1} (\sqrt{m}))}{m}$$ where $n$ is an arbitrary positive integer. Then $$\frac{km^2(m+1)(km+1)}{4}$$ $$= \frac{\sinh^2 (n \sinh^{-1} (\sqrt{m})) \cdot m \cdot (m+1) \cdot (\sinh^2 (n \sinh^{-1} (\sqrt{m})) +1)}{4}$$ evaluates to a positive integer.
I have tried to further simplify $f(m, km)$ given in the conjecture, and I finally get the following result: $$f(m, km)=\frac{m^2a^4+ma^4+m^2a^2+ma^2}{4}$$ for which $a=\sinh (n \sinh^{-1} (\sqrt{m}))$.
Triangular Numbers
By looking at the formula for $f(m, m)$, it equals to the product of two triangular numbers. This may be a new way to solve this problem, but I do not have any progress on this.
