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Claim. The note A Pell classification for k = 2 square representations in Erdős problem #686 is by Nat Sothanaphan and says it was generated in a near-autonomous process by GPT-5.4 Thinking. It claims that $a^2=\frac{(m+1)(m+2)}{(n+1)(n+2)}$ has a solution with $n\ge0$ and $m\ge n+2$ exactly when $a=A_r(y)$ for an odd $y\ge3$ and an $r\ge2$, where $(y+\sqrt{y^2-1})^r=X_r(y)+A_r(y)\sqrt{y^2-1}$. Then $n=(y-3)/2$ and $m=(X_r(y)-3)/2$ give a representation. Thus $A_2(y)=2y$, $A_3(y)=4y^2-1$ and $A_4(y)=4y(2y^2-1)$. For example, $36=A_2(3)^2$ and $1225=A_3(3)^2$ are representable.

Submission note. Posted to the site's forum by Nat Sothanaphan on 8 March 2026:

The $k=2$ case is now settled by GPT-5.4 Thinking with near-autonomous process in this note. The solution uses Pell theory.

Combined with the previously known non-square $N$ case, the classification of representable $N$ for $k=2$ is as follows.

1. $N$ is non-square.
2. $N=a^2$, where $a = A_r(y)$ for some odd integer $y \ge 3$ and integer $r \ge 2$, and $A_r(y)$ is defined by $(y + \sqrt{y^2-1})^r = X_r(y) + A_r(y) \sqrt{y^2-1}$.

The first few polynomials $A_r(y)$ are: $A_2(y) = 2y$, $A_3(y) = 4y^2-1$, $A_4(y) = 4y(2y^2-1)$, coinciding with known parametrizations.

Covers. The question of Problem 686 for each square $A_r(y)^2$, answered yes with $k=2$. The claim that no other square is representable with $k=2$ settles no instance, since larger $k$ remain. The non-square case, which the posting takes from the forum, is recorded on its own page.

Depends on. No other wiki page; the square case rests on the note alone.

Acceptance. None recorded: there is no review, publication or formalization, and the site labels the problem OPEN.