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Let all variables be positive integers. Consider the forms,

$$a^2+b^2=(b+1)^2\tag1$$ $$a^3+b^3+c^3=(c+1)^3\tag2$$

The smallest solutions are the well-known,

$$3^2+4^2=5^2$$ $$3^3+4^3+5^3=6^3$$

All solutions of Eq.1 are covered by only one polynomial parameterization. On the other hand, not all solutions of Eq.2 are covered by its three known parameterizations.


I. Family 1

$$(3 n^2)^3 + (6 n^2 - 3n + 1)^3 + (9 n^3 - 6n^2 + 3n - 1)^3 = (9 n^3 - 6n^2 + 3n)^3$$

$$3^3 + 4^3 + 5^3 = 6^3\\ 12^3 + 19^3 + 53^3 = 54^3\\ 27^3 + 46^3 + 197^3 = 198^3$$

Note: $F(n) = (6, 54, 198)/6 = 1, 9, 33,\dots$ is A005920, or the "tricapped prism numbers".


II. Family 2

$$(3 n^2)^3 + (6 n^2 + 3n + 1)^3 + (9 n^3 + 6n^2 + 3n)^3 = (9 n^3 + 6n^2 + 3n + 1)^3$$

$$3^3 + 10^3 + 18^3 = 19^3\\ 12^3 + 31^3 + 102^3 = 103^3\\ 27^3 + 64^3 + 306^3 = 307^3$$

Note: $F(n) = (18, 102, 306)/3 = 6, 34, 102,\dots$ is A067389.


III. Family 3

$$(n)^3 + (3n^2 + 2n + 1)^3 + (3n^3 + 3n^2 + 2n)^3 = (3n^3 + 3n^2 + 2n + 1)^3$$

$$1^3 + 6^3 + 8^3 = 9^3\\ 2^3 + 17^3 + 40^3 = 41^3\\ 3^3 + 34^3 + 114^3 = 115^3$$

Note: $F(n) = 3n^3 + 3n^2 + 2n = 8, 40, 114,\dots$ is A143943, or "the Wiener index of a chain of $n$ squares".


Update: (Jan 22, four months later)

Courtesy of an insight by Adam Bailey in this post, given $x^3+y^3+z^3=(z+1)^3$ we now realize that $(x,y)$ of Family 2 and 3 are just lattice points on the same ellipse. Do the substitution, $z = (k + 1)(x - 1) + k y\,$ where $k = \dfrac{3n^2+1}{3n}$ and, after removing a trivial factor, we get the ellipse,

$$3n^2(x^2 - xy + y^2) - (1 + 6n + 12n^2 + 18n^3 + 9n^4)x - (1 + 3n^2 + 9n^4)y + (1 + 3n + 9n^2 + 9n^3 + 9n^4) = 0$$

where two lattice points are $(x_1, y_1)$ and $(x_2, y_2)$ from Family 2 and 3, respectively. For example, let $n=3$ so,

$$27(x^2 - x y + y^2) - 1342x - 757y + 1063 = 0$$

which for this $n$ has only 2 lattice points,

enter image description here

thus we can use these coordinates,

$$27^3 + 64^3 + 306^3 = 307^3\;\\ 3^3 + 34^3 + 114^3 = 115^3$$

and so on for infinitely other $n$.


IV. Unknown Families?

$$14^3 + 23 ^3 + 70^3 = 71^3\\ 21^3 + 46^3 + 188^3 = 189^3\\ 16^3 + 51^3 + 213^3 = 214^3\\ \;9^3 \,+\, 58^3 + 255^3 = 256^3\\ 15^3 + 64^3 + 297^3 = 298^3$$

These and the families above are the primitive solutions with $c<307$.


V. Similar equalities

By reverse-engineering those identities,

$$(an^2+bn+c)^3+(dn^2+en+f)^3+(pn^3+qn^2+r)^3=(pn^3+qn^2+rn+s)^3$$

collecting powers of $n$, equating everything to zero, it seems those three are the only quadratic-cubic identities. Turns out there is also the form $a^3+b^3+c^3 = (c\color{blue}{+3})^3$,

$$(m^2)^3 + (2 m^2 + 3m + 3)^3 + (m^3 + 2m^2 + 3m)^3 = (m^3 + 2m^2 + 3m \color{blue}{+3})^3$$

$$(m^2)^3 + (2 m^2 - 3m + 3)^3 + (m^3 - 2m^2 + 3m \color{blue}{-3})^3 = (m^3 - 2m^2 + 3m)^3$$

which, for $m=3n$, can be factored out into $a^3+b^3+c^3 = (c+1)^3$.


VI. Question

Just like $a^3\pm b^3 \pm c^3 = 1$ has infinitely many polynomial parameterizations (where degrees are getting higher and higher), does $a^3+b^3+c^3 = (c+1)^3$ also have other families with degree higher than the cubic polynomials in this post?

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    $\begingroup$ As I suspected, there may be higher degree versions, but one term may have the wrong sign. For example, $$-( m^3 - m^2 + 3m)^3 + (m^3 + 2m^2 + 3)^3 + (m^4 + 2m^2)^3 = (m^4 + 2m^2 + 3)^3$$ If $m=3n$, this becomes $-a^3+b^3+c^3 = (c+1)^3.$ So close! $\endgroup$ Sep 27, 2023 at 14:19
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    $\begingroup$ This is a little above my current understanding, but does this link help? people.math.harvard.edu/~elkies/4cubes.html I coded a simple VB6 program and found $171$ solutions with $c<26755$ Did you miss (22,27,255,256)? Do you want me to post my results and in what order? $\endgroup$
    – Old Peter
    Sep 29, 2023 at 18:50
  • $\begingroup$ @OldPeter Thanks! Yes, I missed that (though it should be 57, not 27). Kindly post it as a partial answer and maybe someone can find some patterns. For example, we now have two pairs of solutions with the same $c+1$, $$22^3 + 57^3 + 255^3 = 256^3\\ 9^3 + 58^3 + 255^3 = 256^3\\ 51^3 + 82^3 + 477^3 = 478^3\\ 64^3 + 75^3 + 477^3 = 478^3$$ I've seen such behavior before and usually involves a parameterization with two terms with only even powers like $$(9t^3 - 1)^3 + (9t^4 - 3t)^3 + 1^3 = (9t^4)^3$$ hence the two terms are immune to sign changes of $t$. Pls arrange the data by increasing $c+1$. $\endgroup$ Sep 30, 2023 at 3:10
  • $\begingroup$ @TitoPiezasIII, I have found a simple way to generate solutions in different families. Following your advice to Old Peter, I will post it as a partial answer ( or a long comment). $\endgroup$
    – user25406
    Sep 30, 2023 at 10:58
  • $\begingroup$ @TitoPiezasIII The following $c$ values give solution pairs $255, 477, 1331, 8898, 13088, 14526, 18755, 20154, 22968$. I’ll attempt to post the list later. $\endgroup$
    – Old Peter
    Sep 30, 2023 at 15:25

5 Answers 5

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I’ve produced $4405$ solutions to $a^3+b^3+c^3=(c+1)^3$.

Of these $293$ are of Family 1, $(3n^2,6n^2−3n+1,9n^3−6n^2+3n−1)$

Of these $292$ are of Family 2, $(3n^2,6n^2+3n+1,9n^3+6n^2+3n)$

Of these $216$ are of Family 3, $(n,3n^2+2n+1,3n^3+3n^2+2n)$

That leaves $3604$ to be classified. I’ve moved the $128$ cases, where there are pairs of solutions with equal $c$, to a new question, $a^3+b^3+c^3=(c+1)^3=A^3+B^3+c^3$ solutions. where I found

I’ve found $8$ cases of a new Family 4,

$$a=9n^3$$

$$b=27n^4+18n^3+9n^2+3n+1$$

$$c=81n^6+81n^5+54n^4+27n^3+9n^2+3n$$

And $8$ cases of a new Family 5, $(A,B,c)$

$$A=9n^3+9n^2+3n+1$$

$$B=27n^4+18n^3+9n^2+3n$$

$$c=81n^6+81n^5+54n^4+27n^3+9n^2+3n$$

So, $a^3+b^3+c^3 = (c+1)^3$ DOES also have other families with degree higher than the cubic polynomials in this post.

However, this just scratches the surface, there is plenty of life left in this question.

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    $\begingroup$ +1 Beautiful! I especially love the simplicity of the first term of family 4, and how families 4 and 5 have the same third term $c$. $\endgroup$ Nov 7, 2023 at 1:28
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    $\begingroup$ @TitoPiezasIII, that is most kind of you to say so, it inspires me to continue. $\endgroup$
    – Old Peter
    Nov 7, 2023 at 11:54
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    $\begingroup$ There seems to be a connection between Family 2 and 4. Note that Family 2 is, $$\color{blue}{(3n^2)}^3 + (6n^2 + 3n + 1)^3 - \color{blue}{(9n^3 + 6n^2 + 3n + 1)}^3 = -z^3$$ while Family 4 is, $$\big(3n\color{blue}{(3n^2)}\big)^3 + \big((3n)\color{blue}{(9n^3 + 6n^2 + 3n + 1)} + 1\big)^3 + c^3 = (c + 1)^3$$ though it could be just coincidence. $\endgroup$ Nov 7, 2023 at 12:12
  • $\begingroup$ Of the the 4405 solutions to $a^3+b^3+c^3=(c+1)^3$, your data says that about 600 have the $a$-variable of form either $3n^2$ or $9n^3$. It might be informative to check other solutions which have the $a$-variable as powers or near-powers. $\endgroup$ Nov 7, 2023 at 12:19
  • $\begingroup$ What is $-z^3$ please? $\endgroup$
    – Old Peter
    Nov 7, 2023 at 14:12
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Here is a further parametric solution, call it Family 6:

$\qquad a=27n^4+54n^3+45n^2+19n+3$

$\qquad b=54n^4+81n^3+63n^2+23n+4$

$\qquad c=243n^6+567n^5+675n^4+468n^3+201n^2+49n+5$

$\qquad c+1 = 243n^6+567n^5+675n^4+468n^3+201n^2+49n+6$

Alternatively,

$\qquad a= (3n + 1)(9n^3 + 15n^2 + 10n + 3)$

$\qquad b= (3n + 1)^2(6n^2 + 5n + 3) + 1$

$\qquad c= (3n + 1)^2(9n^2 + 9n + 5)(3n^2 + 2n + 1) $

$\qquad c+1 = (3n + 1)^2(9n^2 + 9n + 5)(3n^2 + 2n + 1) + 1$

It intersects with Families 1 and 3 when $n=0$ and $n=-1$ respectively.


I found this solution by starting from the general parametric integer solution of $x^3+y^3+z^3=t^3$ due to Choudhry (1) and searching for a condition on the parameters that would ensure the necessary difference of $1$. Choudhry’s solution (with a change of letters) is:

$\qquad sx=r(-p^3-q^3+r^3)$

$\qquad sy=-p^4+2p^3q-3p^2q^2+2pq^3-q^4+(p+q)r^3$

$\qquad sz=p^4-2p^3q+3p^2q^2-2pq^3+q^4+(2p-q)r^3$

$\qquad st=r(p^3+(p-q)^3+r^3)$

To limit the search to manageable proportions, I set $r=1$ and looked for solutions with $x,y < 0$, $z,t > 0$ and $-y+1=z$. The point of the latter is that $x^3+y^3+(-y+1)^3=t^3$ can be rearranged as $(-x)^3+t^3+(-y)^3=(-y+1)^3$ with all terms positive, while $sy+sz=3pr^3=s(y-y+1)=s$, so (with $r=1$) the required common factor $s$ must equal simply $3p$. Aiming for $t=z+1$ looks more complicated.

For $3p$ to be a common factor it suffices that $3|p$ and $3p|(q^3-1)$. This can be seen by re-writing Choudhry’s solution (with $r=1$) as below:

$\qquad sx=-p(p^2)-(q^3-1)$

$\qquad sy= -p^2(p^2-2pq+3q^2)+3pq^3-(p+q)(q^3-1)$

$\qquad sz=p^2(p^2-2pq+3q^2)-(2p+q)(q^3-1)$

$\qquad st=2p(p^2)+3pq(-p+q)-(q^3-1)$

To satisfy $3|q^3-1$ it suffices to put $q=3n+1$ implying $q^3-1=27n^3+27n^2+9n$. Putting $p=(27n^3+27n^2+9n)/3n=9n^2+9n+3$ we then have both $3|p$ and $3p|q^3-1$.

It is then straightforward to substitute for $p$ and $q$ in the above rewritten solution and make the identification $(a,b,c,c+1)=(-x,t,-y,z)$ yielding the parametric form for Family 6 above.

This method can perhaps be adapted to find further families of solutions.

Reference:

  1. Choudhry A (1998) On Equal Sums of Cubes Rocky Mountain Journal of Mathematics Vol 28(4) p 1256
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  • $\begingroup$ An excellent post! I confirm that for $$n= 1$ to $6$ this matches my results. $\endgroup$
    – Old Peter
    Nov 8, 2023 at 19:41
  • $\begingroup$ +1 Great! With 6 families already, there could be infinitely many. And if there is, the polynomials don’t grow as fast as the ones for $x^3+y^3+z^3 =1$, so if there is a recursion, it would be simpler. $\endgroup$ Nov 9, 2023 at 0:19
  • $\begingroup$ @Adam There are infinitely many polynomial solutions of form cubic/quartic, distinct from your quartic/quartic. See my third answer to Peter's post. $\endgroup$ Nov 18, 2023 at 4:06
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Partial answer

First, a prime must be chosen. We pick $7$. Then $7$ is decomposed the following way:
$7=1+6=2+5=3+4$
This decomposition is unique to this prime. In this case we need to consider the $3$ cases and find possible soltions. I chose to start with $3+4$ because it's a solution in the first family.
We are looking for a solution of the form:

$3^3 + 4^3 = a^3 - b^3= (a-b)(a^2 +ab + b^2)$

Since $a-b=1$ or $a=b+1$, the equation becomes: $3^3 + 4^3 =a^2 + ab +b^2$ which after substitution of $a$ by $b+1$ becomes: $3^4 + 4^3 =91=3b^2 + 3b +1$, a simple quadratic equation.

$3b^2 + 3b +1= 91$ becomes $3b^2 + 3b +90 =0$ or $b^2 +b - 30=0$, basically a triangular number in disguise.
The solutions are $b=5$ and $b=-6$. Using this values we can write the solution involving both $a$ andd $b$ as:

$3^3 + 4^3 + 5^3 = 6^3$ which is the first equation of the first family.

Next we consider the decomposition $7=2+5$. The equation in $b$ is given by:
$b^2 + b -44 = 0$. The determinant $\Delta$ is not a square in this case so there will not be an integer solution for $(a,b)$.

Last, the decomposition $1+6$ is considered. Following the same steps above, we end up with the quadratic: $b^2 + b - 216=0$ whose solutions are $8$ and $-9$. The equation becomes: $1^3 +6^3 + 8^3 =9^3$ which is the first solution of the third family.

We can consider any prime and do a decompisition and find the corresponding cubic forms. For example, if we consider the prime $37$, we will end up with the following decompositions:
$37=3+34$ and $14+23$.
These decompositions will give us the third and first solution of the third family and fourth family.
If we consider the prime $67$, we will end up, among many solutions, with the $3$ following solutions of the forth family namely:

$21 + 46=67$
$16 + 51=67$
$9 + 58 =67$

This method doesn't distinguish between the families because it uses a quadratic equation to find the solutions. Here again, the triangular numbers keep showing up when they are the least expected.

In fact the method can be generalized to composites to solve $a^3 + b^3 +c^3= d^3$ with $d=c+2$, or $d=c+4$...We will give only one example.

The equation $a^3-b^3=(a-b)(a^2 +ab + b^2)$ becomes:

$2[(b+2)^2 +(b(b+2) +b^2]=6b^2 +12b + 8$

If we take the first equation of the first family, we get:

$3^3 + 5^3 = 152 = 6b^2 + 12b +8$ which becomes:

$6b^2 + 12b -144 =0$

a quadratic with the solutions $b=-6,4$. This equation corresponds to the decomposition of $8$ into $3+5$. Other decompositions of 8 may or may not admit integer solutions.

So again, we see that the method can handle other cases than $d=c+1$ and can use primes or composites as a starting point. The choice of $8$ in the previous example was used just to show that the method works. In general, the starting point is always an integer and its decomposition into a series of two numbers $N=x_{1} +x_{2}$ then using a quadratic derived from the condition $d = c+i$, with $i=1,2,3...$

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    $\begingroup$ A necessary (but not sufficient) condition for this to work is that the chosen prime is of the form $6n+1$. This is a consequence of the fact that if $a^3+b^3+c^3=d^3$ then $6 | (a+b+c-d)$. $\endgroup$ Sep 30, 2023 at 14:22
  • $\begingroup$ @AdamBailey, in fact the method can be generalized to composites and to having c+2=d, c+3=d, ... For example, we can see that $6^3 - 4^3 =152 =8*19$. I will add this to the answer when I have a bit more time. $\endgroup$
    – user25406
    Sep 30, 2023 at 18:06
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    $\begingroup$ @user25406. I like this method and intent to study it further when I can find the time. $\endgroup$
    – Old Peter
    Sep 30, 2023 at 19:06
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    $\begingroup$ @user25406: Yes, it has also occurred to me to express the equation as $a^3+b^3 = (c+1)^3-c^3$ so both sides factor. But I find it hard to generate polynomial solutions from this approach. $\endgroup$ Oct 1, 2023 at 10:00
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    $\begingroup$ @user25406 As an example of the usefulness of polynomial solutions, if you have two simultaneous quadratics to be made squares, you need a polynomial solution for one, then the problem reduces to solving an elliptic curve as in this post. As an example on how to generate them using Pell equations, like polynomial solutions to $x^3+y^3+z^3=1$, see this post. And I also suggest Dickson's book "History of the Theory of Numbers" for a more enlightening exposition on such matters. $\endgroup$ Oct 1, 2023 at 12:03
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Numerical solutions of $A^3+B^3+C^3=(C+1)^3$ arranged by $C$. There are nine pairs curiously with the same $C$, namely (255, 477, 1331, 8898, 13088, 14526, 18755, 20154, 22968), a sample of which,

\begin{align} 51^3+82^3+477^3 &= 478^3\\ 4^9+75^3+477^3 &= 478^3\\[4pt] 9^9 + 502^3 + 13088^3 &= 13089^3\\ 16^3 + 801^3 + 13088^3 &= 13089^3\\[4pt] 7^9 + 840^3 + 14526^3 &= 14527^3\\ 294^3 + 847^3 + 14526^3 &= 14527^3\\[4pt] 10^9 + 381^3 + 18755^3 &= 18756^3\\ 1018^3 + 69^3 + 18755^3 &= 18756^3 \end{align}

(3,4,5,6) (1,6,8,9) (3,10,18,19) (2,17,40,41) (12,19,53,54) (14,23,70,71) (12,31,102,103) (3,34,114,115) (21,46,188,189) (27,46,197,198) (16,51,213,214) (4,57,248,249) (22,57,255,256) (9,58,255,256) (15,64,297,298) (27,64,306,307) (58,75,453,454) (5,86,460,461) (64,75,477,478) (51,82,477,478) (20,89,487,488) (48,85,491,492) (57,82,495,496) (61,90,564,565) (66,97,632,633) (58,105,671,672) (48,109,684,685) (6,121,768,769) (91,120,909,910) (23,140,958,959) (75,136,989,990) (7,162,1190,1191) (75,166,1290,1291) (44,173,1324,1325) (136,141,1331,1332) (37,174,1331,1332) (76,171,1346,1347) (8,209,1744,1745) (108,199,1745,1746) (96,205,1779,1780) (149,212,2068,2069) (187,192,2130,2131) (108,235,2178,2179) (148,225,2208,2209) (66,247,2262,2263) (9,262,2448,2449) (118,279,2790,2791) (147,274,2813,2814) (10,321,3320,3321) (52,321,3327,3328) (147,316,3402,3403) (22,327,3414,3415) (248,275,3466,3467) (18,349,3764,3765) (48,355,3866,3867) (131,356,3973,3974) (67,372,4154,4155) (28,375,4193,4194) (192,361,4247,4248) (11,386,4378,4379) (270,343,4473,4474) (241,366,4583,4584) (146,395,4645,4646) (192,409,5016,5017) (232,399,5033,5034) (145,426,5175,5176) (335,368,5398,5399) (131,446,5506,5507) (12,457,5640,5641) (243,460,6101,6102) (165,478,6156,6157) (49,522,6888,6889) (243,514,7074,7075) (13,534,7124,7125) (305,506,7255,7256) (24,547,7386,7387) (225,538,7463,7464) (426,451,7506,7507) (376,495,7625,7626) (205,558,7796,7797) (246,571,8186,8187) (456,475,8205,8206) (97,594,8376,8377) (300,571,8429,8430) (448,519,8750,8751) (421,540,8795,8796) (14,617,8848,8849) (115,618,8898,8899) (72,619,8898,8899) (261,616,9156,9157) (314,605,9172,9173) (434,557,9211,9212) (90,643,9426,9427) (139,648,9570,9571) (300,631,9630,9631) (361,636,10071,10072) (114,673,10104,10105) (282,667,10314,10315) (100,693,10548,10549) (485,614,10732,10733) (15,706,10830,10831) (549,598,11244,11245) (363,694,11285,11286) (542,611,11362,11363) (513,634,11399,11400) (564,595,11402,11403) (493,654,11540,11541) (216,775,12590,12591) (363,760,12738,12739) (587,674,13018,13019) (502,729,13088,13089) (16,801,13088,13089) (190,801,13175,13176) (289,798,13320,13321) (264,823,13854,13855) (319,822,13998,13999) (333,820,14003,14004) (288,835,14213,14214) (627,712,14229,14230) (343,840,14526,14527) (294,847,14526,14527) (432,829,14723,14724) (318,865,15048,15049) (17,902,15640,15641) (621,802,15867,15868) (415,888,16038,16039) (432,901,16452,16453) (390,913,16536,16537) (467,902,16690,16691) (82,951,16937,16938) (216,997,18267,18268) (408,985,18471,18472) (18,1009,18504,18505) (547,954,18546,18547) (381,1000,18755,18756) (69,1018,18755,18756) (507,976,18797,18798) (522,979,18978,18979) (655,948,19433,19434) (619,978,19770,19771) (589,990,19787,19788) (643,984,20154,20155) (594,1003,20154,20155) (808,885,20171,20172) (305,1064,20272,20273) (618,1009,20520,20521) (673,990,20616,20617) (507,1054,20826,20827) (58,1095,20921,20922) (648,1015,20958,20959) (367,1086,21057,21058) (774,967,21353,21354) (45,1114,21467,21468) (19,1122,21698,21699) (496,1095,21870,21871) (823,990,22566,22567) (7,1152,22574,22575) (798,1015,22758,22759) (847,984,22806,22807) (865,978,22968,22969) (822,1009,22968,22969) (840,1003,23106,23107) (913,954,23304,23305) (888,979,23370,23371) (588,1135,23561,23562) (911,980,23785,23786) (535,1158,23846,23847) (411,1192,24242,24243) (20,1241,25240,25241) (370,1251,25874,25875) (588,1219,25914,25915)

If anybody could remind me how to produce LF I'll do it.

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  • $\begingroup$ OldPeter: I removed the latex format so it loads faster. I hope you don't mind. These are the 171 smallest primitive solutions? $\endgroup$ Oct 1, 2023 at 9:31
  • $\begingroup$ @TitoPiezasIII, thanks for your help, it’s a long time since I used this markup language. An offshoot question has occurred to me; are you ok if I post it? $\endgroup$
    – Old Peter
    Oct 1, 2023 at 13:50
  • $\begingroup$ Sure, feel free to post your question. Perhaps put a link to this post though. $\endgroup$ Oct 1, 2023 at 15:16
  • $\begingroup$ I think I found a possible explanation for when we get pairs with the same $c$. I am taking the equations $22^3 +57^3 + 255^3 = 256^3$ and $9^3 + 58^3 +255^3 = 256^3$. Using the quadratic equation from my answer below, we end up with the equation: $b^2 + b - 65280 = 0$ with solutions $b=255,-256$ confirming that the decomposition of $22 + 57 = 79$ and $9+58 = 67$ produce the same $c$ and $c+1$. Now $256^3 - 255^3 = 195841$ with $195841 = 37* 67*79$. Notice that both $67$ and $79$ are factor of $195841$. I think that is the reason why we get a pair with the same $c$. (cont) $\endgroup$
    – user25406
    Oct 1, 2023 at 15:48
  • $\begingroup$ (Cont) - Though I have not checked other solutions (because I do everything by hand with a small calculator). I suspect a code can check quickly if other solutions will also show that common factors of $(c+1)^3 -c^3$ is the reason ( sorry I can't code). $\endgroup$
    – user25406
    Oct 1, 2023 at 15:53
0
$\begingroup$

We explore solving the equation for a given value of $c$.

Start by expanding the binomial power $(c+1)^3$ and rearranging:

$a^3+b^3=3c^2+3c+1$

Let $3c^2+3c+1=pq$ for any divisor $p$ of $3c^2+3c+1$. Then we have

$a+b=p$

$a^2-ab+b^2=(a+b)^2-3ab=q$

and thus

$ab=(p^2-3q)/3.$

Factors of $3c^2+3c+1$ are all $\equiv1\bmod3$, so $ab$ will properly be an integer. But $p$ must be chosen large enough to render this product positive yet small enough for the discriminant

$\Delta=(a+b)^2-4ab=(a-b)^2$

to be nonnegative. Thus

$(3c^2+3c+1)^{1/3}\le p\le[(3/4)(3c^2+3c+1)]^{1/2}.$

Solutions are then obtained for those cases where $\Delta$ is a square and thus $a-b$ may be rendered a whole number.

Let us drop in $c=70$ as an example. Then $3c^2+3c+1=14911=13×31×37$, and we have potenially eight choices for $p$ (and thus $q$). But $p=13$ is too small to render a positive value for $ab$ and $p=13×31$ is too large giving a negative $\Delta$. Therefore the only possible solutions are given by $p=31$ or $p=37$.

For $p=31$ we calculate:

$q=13×37=481$

$ab=(31^2-481)/3=160$

$\Delta=31^2-(4×160)=321.$

The discriminant, divisible only by $3$ and not by $9$, fails to be a square.

For $p=37$, however:

$q=13×31=403$

$ab=(37^2-403)/3=322$

$\Delta=37^2-(4×322)=81=9^2.$

So we accept $a+b=p=37$ and $a-b=9$, giving the quoted solution $23^3+14^3+70^3=71^3$.

$\endgroup$
7
  • $\begingroup$ Your answer is the same as my partial answer above. We end up having to solve a quadratic equation in $c$. $\endgroup$
    – user25406
    Nov 8, 2023 at 19:04
  • $\begingroup$ We have a slight difference in sequencing; you sort solutions accodring to $a+b$ and I sort them according to $c$. $\endgroup$ Nov 8, 2023 at 20:50
  • $\begingroup$ not really. I used $a+b$ for the formula $a^3 -b^3$ not the $a,b$ of the expression $a^3 +b^3 +c^3 =(c+1)^3$. I know, it's a bit confusing but the formula on the web provided $a^3-b^3=...$ I didn't bother changing it to $(c+1)^3 -c^3$. $\endgroup$
    – user25406
    Nov 8, 2023 at 20:56
  • $\begingroup$ Here another answer of mine with the "same sequencing" as yours on mathoverflow last October. mathoverflow.net/questions/413031/… $\endgroup$
    – user25406
    Nov 9, 2023 at 11:10
  • 1
    $\begingroup$ Mistyped, sorry. $\endgroup$ Nov 9, 2023 at 13:51

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