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I am interested in all solutions in odd positive integers d, e, k, with d<=k and e<=k of the equation d^2 + (k-1)e^2 = k(k^2 + 2).

This came up during work on dominating sets of the queen's graph (finding for each n the minimum number of queens one can place on an n-by-n board so that every square is either occupied or covered by a queen).

For each odd positive k, there is at most one odd positive e giving a solution (k, e, d).

I can see see that there are infinite sequences of solutions, related to Pell-type equations of the form X^2 - 3Y^2 = constant.

The density of solutions appears to decrease as k increases.

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    $\begingroup$ This equation seems to have no solutions: $0<d\leq k$ and $0<e\leq k$, so $d^2+(k-1)e^2\leq k^2+(k-1)k^2<k(k^2+2)$ $\endgroup$ – Alexander Kalmynin Jun 7 at 22:05
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    $\begingroup$ I am guessing that the question should be corrected to something like $d^2+(k-100)e^2=100 k(k^2+2) $ so that solutions are allowed. One quick way to get bounds for the number of solutions of this kind of equations is to reduce modulo $k$. Namely, one must have $d^2-100 e^2\equiv 0 \mod{k} , $ hence, for every $d \leq k $ there are at most $O(k^\epsilon)$ values for $e$. Hence, the bound is $O(k^{1+\epsilon})$, for every fixed $\epsilon>0$. $\endgroup$ – Dr. Pi Jun 9 at 11:13

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