Consider the sums of cubes of three distinct positive integers in arithmetic progression, i.e. $a^3 + (a+b)^3 + (a+2b)^3$ where $a$ and $b$ are positive integers.
$$5643 = 1^3 + 9^3 + 17^3 = 6^3 + 11^3 + 16^3$$ is the smallest positive integer that has two such representations.
$$255816 = 8^3 + 34^3 + 60^3 = 18^3 + 38^3 + 58^3 = 43^3 + 44^3 + 45^3$$ is the smallest that has three such representations.
Is there a positive integer with four or more such representations, and if so what is the smallest?
The sequence of positive integers with one or more representations is OEIS sequence A306213. Those with two or more is A359055.
If representations where the three cubes do not need to be positive are allowed, then I know of examples with up to $11$ representations. Thus
$$ \eqalign{42878854656 &= (-17232)^3 + 24^3 + 17280^3\cr
&= (-6900)^3 + 144^3 + 7188^3\cr
&= (-6704)^3 + 152^3 + 7008^3\cr
&= (-3132)^3 + 528^3 + 4188^3\cr
&= (-1812)^3 + 912^3 + 3636^3\cr
&= (-1032)^3 + 1224^3 + 3480^3\cr
&= (-292)^3 + 1552^3 + 3396^3\cr
&= (-24)^3 + 1672^3 + 3368^3\cr
&= 1038^3 + 2112^3 + 3186^3\cr
&= 1635^3 + 2304^3 + 2973^3\cr
&= 1728^3 + 2328^3 + 2928^3\cr
}$$
Best Answer
I think there is no point in trying to guess any number. First you need to write a parametrization of the solutions of the equation.
$$(A-B)^3+A^3+(A+B)^3=(C-Q)^3+C^3+(C+Q)^3$$
$$A=4(24a^2-8ab+b^2)$$
$$B=177a^2-59ab+6b^2$$
$$C=4(33a^2-10ab+b^2)$$
$$Q=132a^2-49ab+6b^2$$
Then you can substitute it into the formula and then play with the coefficients. If this fails, then you will have to solve a system of nonlinear equations. There will be very cumbersome calculations.
There was a question. When there are solutions $ A−B>0 $ and $ C−Q>0 $
Each attempt at a solution gives a new formula for some reason. It didn't fit.
$$A=2(36a^2-4ab+b^2)$$
$$B=64a^2-ab+3b^2$$
$$C=2(32a^2+b^2)$$
$$Q=74a^2-11ab+3b^2$$
The following formula gives the necessary solutions.
$$A=528a^2-40ab+b^2$$
$$B=64a^2-25ab+3b^2$$
$$C=128a^2+b^2$$
$$Q=764a^2-95ab+3b^2$$
The general formula should contain 3 more parameters.
At the moment, we have managed to obtain such a formula.
$$A=64k^{4}t^{3}x^{2}+2ty^2$$
$$B=2k(27k^6-18k^{4}t^2+20k^2t^4+8t^6)x^2+(9k^2+2t^2)(3k^2-2t^2)xy+3ky^2$$
$$C=8tk^2(9k^4-4k^2t^2+4t^4)x^2+8kt(3k^2-2t^2)yx+2ty^2$$
$$Q=64k^3t^4x^2+(3k^2-2t^2)^2xy+3ky^2$$