$a^3+b^3+c^3=(c+1)^3=A^3+B^3+c^3$ solutions.

Question

(Note: Positive integers throughout, except where specified.)

I. Parameterization

Following on from this excellent question from Tito Piezas III, "Sums of Cubes of form $a^3+b^3+c^3=(c+1)^3$". Piezas observed there were pairs that shared the same $c$, hence the same sum $c+1$, for example,

\begin{align} 9^3+58^3+255^3 &=256^3\\ 22^3+57^3+255^3 &=256^3 \end{align}

So I have produced 4405 solutions to $a^3+b^3+c^3=(c+1)^3$ up to $c=3.15\times10^8,$ then extracted $128$ solutions (given in next section) that share the same sum,

$$a^3+b^3+c^3=A^3+B^3+c^3=(c+1)^3$$

For the eight cases where $b-B=1$, I have found a parametric solution

$$a=9n^3$$ $$b=27n^4+18n^3+9n^2+3n+1$$ $$c=81n^6+81n^5+54n^4+27n^3+9n^2+3n$$ $$A=9n^3+9n^2+3n+1$$ $$B=27n^4+18n^3+9n^2+3n$$

Proof: Just expand and equate.

Non-positive $n$ give incorrect results if negative results are moved to change sign. For the majority of cases where $b-B\gt1,$ I have not yet found parametric solutions.

Results:

(n,a,b,c,A,B)

(1,9,58,255,22,57)
(2,72,619,8898,115,618)
(3,243,2764,83925,334,2763)
(4,576,8221,430428,733,8220)
(5,1125,19366,1556115,1366,19365)
(6,1944,39223,4485150,2287,39222)
(7,3087,71464,11030313,3550,71463)
(8,4608,120409,24123480,5209,120408)

For $n=1$ to $8$ these agree with the expected results.

My question:

Can you please find parametric solutions for one or more of the following groups of numerical solutions below to $a^3+b^3+c^3=A^3+B^3+c^3=(c+1)^3$ ?

P.S. Failing that, a link to a reliable sequence formula calculator, or a detailed worked example of converting difference tables to formula manually.

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II. Raw Data

The 128 solutions below are sorted by $b-B = k$ for $k\leq 103$, including 28 with $k> 103.$

128 = 16 + 12 + 8 + 2 + 8 + 2 + 6 + 2 + 4 + 4 + 2 + 4 + 2 + 4 + 4 + 2 + 4 + 4 + 4 + 2 + 4 + 28

Data for $b-B=1$

(22,57,255)
(9,58,255)
(115,618,8898)
(72,619,8898)
(334,2763,83925)
(243,2764,83925)
(733,8220,430428)
(576,8221,430428)
(1366,19365,1556115)
(1125,19366,1556115)
(2287,39222,4485150)
(1944,39223,4485150)
(3550,71463,11030313)
(3087,71464,11030313)
(5209,120408,24123480)
(4608,120409,24123480)

Data for $b-B=7$

(51,82,477)
(64,75,477)
(294,847,14526)
(343,840,14526)
(1302,5971,267762)
(1435,5964,267762)
(2841,16672,1245927)
(3064,16665,1245927)
(6333,48004,6079299)
(6712,47997,6079299)
(10302,91321,15944352)
(10825,91314,15944352)

Data for $b-B=13$

(1956,6631,315726)
(2095,6618,315726)
(2892,11137,684480)
(3073,11124,684480)
(13683,87400,14946453)
(14194,87387,14946453)
(16941,116026,22853211)
(17530,116013,22853211)

Data for $b-B=18$

(69,1018,18755)
(381,1000,18755)

Data for $b-B=19$

(594,1003,20154)
(643,984,20154)
(1719,4258,165609)
(1828,4239,165609)
(11673,54700,7421991)
(12076,54681,7421991)
(18738,102595,19030386)
(19291,102576,19030386)

Data for $b-B=30$

(7739,92810,16328917)
(10739,92780,16328917)

Data for $b-B=31$

(822,1009,22968)
(865,978,22968)
(7401,21256,1826583)
(7648,21225,1826583)
(23403,98914,18079365)
(23944,98883,18079365)

Data for $b-B=33$

(37,174,1331)
(136,141,1331)

Data for $b-B=37$

(2709,4762,206451)
(2818,4725,206451)
(5796,13477,938532)
(5989,13440,938532)

Data for $b-B=43$

(19689,62740,9212247)
(20116,62697,9212247)
(35439,137650,29735697)
(36076,137607,29735697)

Data for $b-B=48$

(46,6441,298448)
(1810,6393,298448)

Data for $b-B=49$

(31359,106888,20429049)
(31918,106839,20429049)
(38523,140740,30794541)
(39166,140691,30794541)

Data for $b-B=57$

(1102,14421,1000067)
(3325,14364,1000067)

Data for $b-B=61$

(8124,14653,1107900)
(8317,14592,1107900)
(14685,32986,3608235)
(14986,32925,3608235)

Data for $b-B=67$

(5898,8629,531564)
(6037,8562,531564)
(26373,68104,10554939)
(26812,68037,10554939)

Data for $b-B=72$

(16,801,13088)
(502,729,13088)

Data for $b-B=73$

(4203,4780,247293)
(4294,4707,247293)
(44064,127975,26965902)
(44671,127902,26965902)

Raw data for $b-B=79$

(6558,8635,555546)
(6691,8556,555546)
(42855,116638,23561961)
(43432,116559,23561961)

Data for $b-B=91$

(19215,35266,4121265)
(19516,35175,4121265)
(31194,68563,10842210)
(31627,68472,10842210)

Data for $b-B=97$ (Second pair added by Piezas)

(11040,15253,1277268)
(11221,15156,1277268)
(60117, 159286, 37677027)
(60790, 159189, 37677027)

Data for $b-B=103$

(16545,25864,2697567)
(16792,25761,2697567)
(55326,136441,30051936)
(55945,136338,30051936)

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More raw data for $k>103$

(39006,72487,12113622) (39439,72360,12113622)

(13380,14689,1361952) (13537,14556,1361952)

(21129,25888,2987895) (21352,25737,2987895)

(47781,78856,14135451)
(48220,78693,14135451)

(71169,133498,30219399)
(71758,133329,30219399)

(54477,80212,15030579)
(54904,80013,15030579)

(32835,35326,5147325)
(33076,35115,5147325)

(62517,80236,15925707)
(62920,79989,15925707)

(67881,78916,16373271)
(68260,78633,16373271)

(68346,72577,15292152)
(68689,72270,15292152)

(201,3166,102863)
(2097,2824,102863)

(3029,12038,768604)
(8021,10790,768604)

(865,20886,1742760)
(15273,17704,1742760)

(8781,72346,11244731)
(47997,64534,11244731)

Note: Despite checking and double-checking, mistakes happen (sorry). Please check before use.

Answer 1

This third answer neatly generalizes both my first and second answers. When I noticed similarities between the polynomial parameterizations I found for ALL primes $p = 6m+1$ with $p<100,$ I knew there had to be a common form. After some effort, I found it. Given,

$$a^3+b^3+C^3 = A^3+B^3+C^3 = (C+1)^3$$

then the two-parameter identity was quite simple as,

\begin{align} a &= \frac{p^3+q^3-1}{3p}\\[4pt] b &= \frac{p^3(q-1)+(q^2+q+1)^2}{3p^2}+\color{blue}p\\[4pt] A &= \frac{p^3+(q+1)^3+1}{3p}\\[4pt] B &= \frac{p^3(q-1)+(q^2+q+1)^2}{3p^2}\\[4pt] C &= \frac{p^6+9p^3q^2+2p^3(q-1)^3+(q^2+q+1)^3}{9p^3} \end{align}

Of course, $b-B = p.\,$ (The new variables $(p,q)$ were chosen for aesthetics.) For example, for $p = 1$, then there is just one choice such that all terms are integers, namely $q = 3n$, and we recover Peter's,

\begin{align} a &=9n^3\\ b &=27n^4+18n^3+9n^2+3n+1\\ A &=9n^3+9n^2+3n+1\\ B &=27n^4+18n^3+9n^2+3n\\ C &=81n^6+81n^5+54n^4+27n^3+9n^2+3n \end{align}

However for $p>1$ and $p = 6m+1 \neq (6e-1)(6f-1)$, then we can have more than one choice depending on the distinct prime factorization of $p$.

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Examples:

For $p=7$, then $q = 3(7n+\alpha)$ for $\alpha$ = (3, 6) which explains the two families in my first answer.

For $p = 7\times13=91$, then $q = 3(91n+\alpha)$ for $\alpha$ = (3, 27, 55, 66) which yields four families.

For $p = 7\times13\times19=1729$, then $q = 3(1729n+\alpha)$ for $\alpha$ = (276, 300, 794, 846, 965, 1340, 1459, 1511) which yields eight families.

And so on. In general, if $p$ is prime, then there are two choices of $\alpha$. If $p$ is not prime, then there may be more. Thus, finding integer polynomial parameterizations to,

$$a^3+b^3+C^3 = A^3+B^3+C^3 = (C+1)^3$$

where $b-B = p$ for some desired $p=6m+1$ can be reduced to the easier divisibility problem,

$$B = \frac{p^3(q-1)+(q^2+q+1)^2}{3p^2}$$

where $q = 3(pn+\alpha)$ and finding appropriate integers $\alpha$ within the small range $\alpha<p$ such that $B$'s denominator vanishes, easily done by Mathematica in seconds.

Answer 2

(New answer, a few hours later.)

As I suspected (see old answer below), while the case $k=1$ involved just one pair of families that solved,

$$a^3+b^3+c^3 = A^3+B^3+c^3 = (c+1)^3$$

the case $k=7$ involved two pairs of families.

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I. Pair 1

\begin{align}a_1 &= 3 (7 n - 2) (21 n^2 - 30n + 12) + 6 = \color{blue}{51,\, 1302,\, 6333},\dots\\[6pt] b_1 &= 3 (441 n^4 - 966 n^3 + 795 n^2 - 275 n + 30) + \color{red}7\\[6pt] c_1 &= 3(9261 n^6 - 30429 n^5 + 41706 n^4 - 29835 n^3 + 11451 n^2 - 2147n + 152)\\[6pt] A_1 &= 3 (7 n - 1) (21 n^2 - 30n + 13) - 8 = \color{blue}{64,\, 1435,\, 6712},\dots\\[6pt] B_1 &= 3 (441 n^4 - 966 n^3 + 795 n^2 - 275 n + 30)\\[6pt] \end{align}

Thus $b_1-B_1 = 7.\,$ (Note that powers of $7$ appear throughout, such as $441 = 21^2,$ and $9261 = 21^3.$) Example for $n=1$,

$$51^3 + (75 + 7)^3 + 477^3 = 478^3\\ 64^3 + (75 + 0)^3 + 477^3 = 478^3$$

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II. Pair 2

\begin{align}a_2 &= 3(7 n + 1) (21 n^2 - 12n + 3) + 6 = \color{blue}{294,\, 2841,\, 10302},\dots\\[6pt] b_2 &= 3 (441 n^4 - 210 n^3 + 39 n^2 + 13n - 3) + \color{red}7\\[6pt] c_2 &= 3 (9261 n^6 - 6615 n^5 + 2016 n^4 + 351 n^3 - 213 n^2 + 31n + 11)\\[6pt] A_2 &= 3 (7 n + 2) (21 n^2 - 12n + 4) - 8 = \color{blue}{343,\, 3064,\, 10825},\dots\\[6pt] B_2 &= 3 (441 n^4 - 210 n^3 + 39 n^2 + 13n - 3)\\[6pt] \end{align}

Likewise, $b_2-B_2 = 7.\,$ Example for $n=1$,

$$294^3 + (840 + 7)^3 + 14526^3 = 14527^3\\ 343^3 + (840 + 0)^3 + 14526^3 = 14527^3$$

Peter's 12 blue numbers were basically all I had to work with, and I had to separate them properly to find these two pairs of families. Plus a lot of guesswork based on the structure of the case $k=1,$ and help from Mathematica.

I assume that, with enough data, parameterizations of $b-B = k\,$ for other $k$ may be found. (They have also been found for $k=13,19,31$. Will be posted later.)

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(Old answer)

I. Case $b-B = 1$

The differences of the first 6 pairs,

$$(22 - 9,\; 115 - 72,\; 334 - 243,\; 733 - 576,\; 1366 - 1125,\; 2287 - 1944)$$

are $(13,\; 43,\; 91,\; 157,\; 241,\; 343).\,$ Their differences divided by 6 are,

$$(5,\; 8,\; 11,\; 14,\; 17)$$

with a constant difference of $\color{blue}{3\times1 = 3}$. They are in arithmetic progression and, testing for the rest of the pairs, it also holds true. Thus $k=1$ has a polynomial parameterization.

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II. Case $b-B = 7$

The differences of the 6 known pairs,

$$(64 - 51,\; 343 - 294,\; 1435 - 1302,\; 3064 - 2841,\; 6712 - 6333,\; 10825 - 10302)$$

are $(13,\; 49,\; 133,\; 223,\; 379,\; 523).\,$ Their differences are also divisible by 6 but do not yield an arithmetic progression. However, if we take differences of every other number,

$$-(13-133,\; 133-379)/6 = (20,\,41)$$ $$-(49-223,\; 223-523)/6 = (29,\,50)$$

then we seem to have a constant difference of $\color{blue}{3\times7 = 21}$. Unfortunately, with limited data, this is hard to test. Maybe more data points are needed. (See new answer above.)

Answer 3

Thanks to Peter's database and a lot of painstaking data-mining and trial-and-error using Mathematica, and guided by the form for cases $k=1$ and $k=7$, we can show there are infinitely many polynomial identities to,

$$a^3+b^3+c^3 = A^3+B^3+c^3 = (c+1)^3$$

Parameterizations can exist if $k=b-B = \color{blue}{9m^2\pm3m+1} = 1,7,13,31,43,\dots$ as well as for others such as $k=19$. Recall from this post that this polynomial is quite important to "real cubic fields".

We can solve the equation as a quadratic in $c$,

$$c = \frac{-3+\sqrt{12a^3+12b^3-3}}6 = \frac{-3+\sqrt{12A^3+12B^3-3}}6$$

Thus we need to find appropriate $(a,b,A,B)$. First, define some variables for the first pair,

$$p = 9m^2,\quad q = 9m^2+1$$

and for the second pair,

$$r = -3m,\quad s = -3m+1$$

Then the two pairs are related by,

$$k = \color{blue}{p+s = q+r} = 9m^2-3m+1 $$

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I. First Pair

Using $p = 9m^2,\;q = 9m^2+1,$

\begin{align} a_1 &= 9k^2n^3 + 9k p n^2 + 3p^2 n + (54m^4 - 9m^3 + 9m^2 - 3m)\\[6pt] b_1 &= B_1+k\\[6pt] A_1 &= 9k^2n^3 + 9k q n^2 + 3q^2 n + (54m^4 - 9m^3 + 18m^2 +1)\\[6pt] B_1 &= 27k^2 n^4 + 18k(18m^2 + 1)n^3 + 9(162m^4 + 18m^2 + 1)n^2 + 3(135m^4 + 18m^3 + 27m^2 + 1)n + (54m^4 + 9m^3 + 9m^2 + 3m) \end{align}

so $b_1-B_1 = k$. Example, let $m=2$, so $k = 31$, and $n=0,1$. Then,

$$822^3 + (978 + 31)^3 + 22968^3 = 22969^3\\ 865^3 \,+ (978 + 0)^3 \,+ 22968^3 = 22969^3\\ 23403^3 + (98883 + 31)^3 + 18079365^3 = 18079366^3\\ 23944^3 + (98883 + 0)^3 \,+\, 18079365^3 = 18079366^3$$

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II. Second Pair

Using $r = -3m,\; s = -3m+1,$

\begin{align} a_2 &= 9k^2n^3 + 9k r n^2 + 3r^2 n + (27m^4 - 18m^3 + 9m^2 - 3m)\\[6pt] b_2 &= B_2+k\\[6pt] A_2 &= 9k^2n^3 + 9k s n^2 + 3s^2 n + (27m^4 - 18m^3 + 9m^2 - 3m + 1)\\[6pt] B_2 &= 27k^2n^4 - 18k(6m - 1)n^3 + 9(18m^2 - 6m + 1)n^2 + 3(27 m^4 - 18m^3 + 9m^2 - 6m + 1)n - 9m^3 \end{align}

Likewise, $b_2-B_2 = k$. Example, let $m=2$, so $k = 31$ again, and $n=1,2$. Then,

$$7401^3 + (21225 + 31)^3 + 1826583^3 = 1826584^3\\ 7648^3 \,+ (21225 + 0)^3 \,+ 1826583^3 = 1826584^3\\ 63030^3 + (370050 + 31)^3 + 130303080^3 = 130303081^3\\ 64081^3 + (370050 + 0)^3 \,+\, 130303080^3 = 130303081^3$$

For $k=31$, Peter found the first six, but not the 7th and 8th as it had $c$ beyond his search range. These four identities "explain" why the number of his solutions come in multiples of 4, and why pairs of first terms are relatively close to each other. However, there is probably another superfamily that contains $k=19, 37,$ and others.