It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.

Accordingly, I have been looking into instances where $c^3|(a^3+b^3)$ for $a,b\in\mathbb{N}$. In other words, solutions to the Diophantine Equation:

$a^3+b^3=dc^3$ where $a,b,$ and $c$ are pairwise co-prime and $a,b,c>0$

Obviously there are some trivial solutions. If $c=1$, for example, $a$ and $b$ can be any integers, and $d$ can be chosen to simply be $a^3+b^3$.

By requiring that $a,b,$ and $c$ are pairwise co-prime and that $c\not=1$, we eliminate the trivial solutions, and what remains is of much more interest.

For $a,b,c,d\le20$, there are 5 solutions:






Under $100$ there are $16$ solutions, as found by Mathematica.

My question about this equation: Has it been studied previously? Are there infinitely many primitive solutions (which it seems like there are)? If so, can they be parametrized?

  • 1
    $\begingroup$ you can formulate the following statement: for any $c \neq 1$, let $\Phi (c)$ be the set of all remainders $\pmod {c^3}$. Then you can ask whether there are two elements $a,b \in \Phi (c)$ such that $a^3 + b^3 \equiv 0 \pmod {c^3}$ or $a^3 - b^3 \equiv 0 \pmod {c^3}$ It is true, for example, for c=2, since $\lambda(8) = 2 $(Carmichael function) and then $5^3 + 3^3 \equiv 5 + 3 \equiv 0 \pmod 8$ $\endgroup$ – Francisco José Letterio Oct 20 '19 at 4:28
  • 1
    $\begingroup$ The obvious cases are $c^3\mid a+b$ and $c^3\mid a^2-ab+b^2$, and the first clearly has many "non-trivial" solutions under your current definition... $\endgroup$ – abiessu Oct 20 '19 at 4:55
  • 3
    $\begingroup$ This is $r^3+s^3=d$, so representing the integer $d$ as a sum of two rational cubes. A lot of work has been done, a lot remains to do. An internet search for sum of two rational cubes should turn up some information – why not carry out this search, and then report back to us on what you have found? $\endgroup$ – Gerry Myerson Oct 20 '19 at 5:02
  • $\begingroup$ From your examples you can generate infinitely many primitive solutions. Viz., if $a=4+27n$ and $b=5+27n$ then $a,b$ are co-prime (because $b-a=1$) and $a^3+b^3\equiv 4^3+5^3\equiv 0 \mod 27.$ $\endgroup$ – DanielWainfleet Oct 21 '19 at 13:18
  • $\begingroup$ For $c>1$ let $1\le a<c^3$ such that $a$ is co-prime to $c^3-a$. E.g. $a=1$ if $c$ is even, or $a=2$ if $c$ is odd. Let $b=c^3-a$. Then $a^3+b^3=(a+b)(a^2-ab+b^2)$ is divisible by $a+b=c^3.$ $\endgroup$ – DanielWainfleet Oct 21 '19 at 13:27

To show that there are infinitely many non-trivial solutions, it suffices to consider solutions of the following form, for $n=0,1,2\dots$

$\qquad a=3$

$\qquad b=24n+5$

$\qquad c=2$

$\qquad d=(3n+1)[9-3(24n+5)+(24n+5)^2]=1728n^3+1080n^2+225n+19$

Note that:

$$a^3+b^3 = (a+b)(a^2-ab+b^2)=(3+24n+5)[9-3(24n+5)+(24n+5)^2]$$

so that:


and also:

$\gcd(a,b)=1$ since $24n+5\equiv2\pmod3$

$\gcd(a,c)=1$ and $\gcd(b,c)=1$ since $a,b$ odd.

| cite | improve this answer | |

It's actually very easy to identify an infinite number of solutions. Suppose that $m^3+1$ is divisible by $3^3=27$, whichnisctrue for $m=8$. Then


is also divisible by $27$. Having $1$ as one of the cubes guarantees relative primarily for all these solutions, but we can reasonably expect infinitely many relatively prime solutions to arise similarly from $m^3+n^3$ for other values of $n$.

| cite | improve this answer | |
  • $\begingroup$ This is not correct as it stands: $19683$ should be $729$ ... $\endgroup$ – Adam Bailey Oct 21 '19 at 10:17
  • $\begingroup$ ... but the idea is good and leads to the identity $(9n+8)^3+1^3=(27n^3+72n^2+64n+19)\times3^3$. $\endgroup$ – Adam Bailey Oct 21 '19 at 10:19
  • $\begingroup$ Edited . . . . . . $\endgroup$ – Oscar Lanzi Oct 21 '19 at 10:20

There are infinitely many parametric solutions for arbitrary c.
We use simple identity below. $$(c^3n+c^3-b)^3+b^3=c^3(n+1)(c^6n^2+2c^6n-3c^3nb+3b^2+c^6-3c^3b)$$ Let $a=c^3n+c^3-b, d=(n+1)(c^6n^2+2c^6n-3c^3nb+3b^2+c^6-3c^3b)$.
Hence we get the parametric solutions of $a^3+b^3=dc^3$.
We show the examples only for $c=2,3,4,5.$

$$(8n+7)^3+1^3 = 2^3(n+1)(64n^2+104n+43)$$

$$(27n+26)^3+1^3 = 3^3(3n+3)(243n^2+459n+217)$$

$$(64n+61)^3+3^3 = 4^3(n+1)(4096n^2+7616n+3547)$$

$$(125n+124)^3+1^3 = 5^3(n+1)(15625n^2+30875n+15253)$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.