I am interested in all solutions in odd positive integers $d$, $e$, $k$, with $d\leq k$ and $e\leq k$ of the equation $d^2 + (k-1)e^2 = k(k^2 + 2)/3$. (I had posted this earlier but left out the division by 3 - I regret any wasted time and energy due to my error.)
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$\begingroup$ Can you expand on the motivation for this? And what you've done with the problem? For a fixed d and e, this should correspond to an elliptic curve for k (aside from a few cases which should be easy to write down in terms of the discriminant). So aside from those cases, there should be finitely many values k for a given d and e, even without your restrictions on the size of k. $\endgroup$– JoshuaZJun 14, 2021 at 23:58
2 Answers
A mindless gp search over all odd $d,e,k < 2048$ takes about 2 minutes to find the following triples $[d,e,k]$:
[1, 1, 1]
[3, 1, 3]
[3, 3, 5]
[7, 5, 9]
[1, 9, 15]
[11, 11, 19]
[27, 15, 27]
[63, 37, 65]
[41, 41, 71]
[115, 67, 117]
[1, 69, 119]
[153, 153, 265]
[363, 209, 363]
[309, 251, 435]
[131, 271, 469]
[45, 293, 507]
[443, 385, 667]
[237, 413, 715]
[645, 505, 875]
[891, 515, 893]
[571, 571, 989]
[177, 823, 1425]
[1615, 933, 1617]
We notice that there are some solutions with $d=e$ and some with $d=k$. Setting $d=e$ or $d=k$ yields the Fermat-Pell equations $3d^2 = k^2+2$ and $3e^2 = k^2-2k$, which have infinitely many solutions. Likewise we detect infinite series of solutions with $d=1$ and $3e^2 = k^2+k+3$, or $d = k-2$ and $3e^2 = k^2 - 2k + 12$. But there also seem to be plenty of other solutions, and there might be no way to account for them all.
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$\begingroup$ I had found those families of Fermat-Pell solutions but wondered if there were more. Also, up to k = 5 x 10^8 there are 187 solutions; I wondered why they seem to be thinning out as k increases. $\endgroup$ Jun 21, 2021 at 22:03
$$k(k^2+2)=3(k-1)e^2+3d^2\tag{1}$$
We derive the quadratic equation for $k$ to get infinitely many integer solutions.
Let $d = k-n$, then we get
$3e^2 = (k^3-3k^2+6kn+2k-3n^2)/(k-1).$
$k^3-3k^2+6kn+2k-3n^2$ is divisible by $k-1$ if $-3n^2+6n=0.$
Hence if $n=0$ and $d=k$ or $n=2$ and $d=k-2$ then we get $3e^2= k^2-2k$ or $3e^2=k^2-2k+12.$
These equations have already been derived by Elkies.
We get Pell's equations as follows.
$3e^2= k^2-2k \implies (k-1)^2-3e^2 = 1$
$3e^2=k^2-2k+12 \implies (k-1)^2-3e^2 = -11$
Example of the case for d=k.
(d,e,k)
(3,1,3)
(27,15,27)
(363,209,363)
(5043,2911,5043)
(70227,40545,70227)
(978123,564719,978123)
(13623483,7865521,13623483)
(189750627,109552575,189750627)
(2642885283,1525870529,2642885283)
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$\begingroup$ My motivation: I wish to find for what 2k-by-2k chessboards there exists a monochromatic set of k queen squares that dominates the board. (With the usual alternating coloring of board squares, this means a set D of k board squares, all of the same color, having the property that every board square is either in D or shares a row, column, or diagonal with a square of D.) $\endgroup$ Jun 17, 2021 at 21:15
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1$\begingroup$ My motivation: I wish to find for what 2k-by-2k chessboards there exists a monochromatic set of k queen squares that dominates the board. (With the usual alternating coloring of board squares, this means a set D of k board squares, all of the same color, having the property that every board square is either in D or shares a row, column, or diagonal with a square of D.) If there is such a set for given k \geq 3, then there exist e and d such that (k,e,d) satisfies the equation I gave. For any odd positive k, the restriction 1 \leq d \leq k means there is at most one value of e that works. $\endgroup$ Jun 17, 2021 at 21:24