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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:

$4^3+5^3=7*3^3$

$2^3+7^3=13*3^3$

$1^3+8^3=19*3^3$

$3^3+5^3=19*2^3$

$1^3+19^3=20*7^3$

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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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:

$$a^3+b^3=(3n+1)[9-3(24n+5)+(24n+5)^2]2^3=dc^3$$

and also:

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

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

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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

$(m+9)^3+1=(m^3+1)+(27m^2+243m+729)$

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 $n$.

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  • $\begingroup$ 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
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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 $c=2,3,4,5.$

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

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

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

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

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