About the diophantine equation $x^3+y^3=z^8$

Question

After solving the exercise

• The equation $x^n+y^n=z^{n+1}$ has infinite solutions in the positive integers.

I realized that this can be easily generalized to

• If $m\equiv 1 \pmod n$, the equation $x^n+y^n = z^m$ has infinite solutions in the positive integers.

Then, for example, $x^3+y^3 = z^7$ has infinite solutions. Then I wondered about the equation $$x^3+y^3=z^8$$

After a numeric search and some simplifications I got two family of solutions:

$$(x,y,z) = (32 n^8,32 n^8,4 n^3), \text{ with } n\ge 1$$ $$(x,y,z) = (18 n^8,9n^8,3n^3), \text{ with } n\ge 1$$

It's clear that from any "primitive" solution $(a,b,c)$ one can get other solutions $(a n^8, b n^8, c n^3)$. The numeric search for $x, y \le 10000$ gives only the two primitive solutions $(32,32,4)$ and $(18,9,3)$

I'm wondering: Are all the solutions of $x^3+y^3=z^8$ covered by these two families?

Answer 1

More solutions can be found. For example $x=3\cdot 28^5$, $y=28^5$, $z=28^2$ when $$x^3+y^3=(28^5)^3(3^3+1^3)=28^{16}=z^8.$$

---

The ones you found can be explained as follows:

• Starting with $1^3+1^3=2$ we see that $(2^a)^3+(2^a)^3=2\cdot2^{3a}=2^{3a+1}$, and it is all about making $3a+1$ divisible by eight, so $a=5$ yields $x=y=32, z=4$.
• Starting with $2^3+1^3=9=3^2$ we similarly see that $(2\cdot3^a)^3+(3^a)^3=3^{3a+2}$, and we need to choose $a$ so that $3a+2$ is divisible by eight. This time $a=2$ gives $x=18, y=9, z=3$.

The same method gives other families. You can start with a random "seed" choice for $x=a$ and $y=b$, and then use $a^3+b^3$ as a "scale parameter" (like $2$, $3$ and $28$ in the three families above) and find solutions with a given $[x:y]=[a:b]$ ratio $$x=a(a^3+b^3)^5,\qquad y=b(a^3+b^3)^5,\qquad z=(a^3+b^3)^2.$$