Primitive Inverse-Cubed Triplets!

Number Theory Level 1

Define a primitive inverse-cubed triplet as any ordered triplet $(x,y,z),$ where $x,y,z\in\mathbb{Z}^+$ and $\gcd{(x,y,z)}=1,$ such that

$\frac{1}{x^3}+\frac{1}{y^3}=\frac{1}{z^3}.$

How many ordered triplets $(x,y,z)$ exist that are primitive inverse-cubed triplets?

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An infinite amount exists A finite amount exists No such

(x,y,z)

exists

2 solutions

Nick Turtle
May 6, 2018

$\begin{aligned}\frac{1}{x^3}+\frac{1}{y^3}&=\frac{1}{z^3}\\ {(yz)}^3+{(xz)}^3&={(xy)}^3\end{aligned}$

By Fermat's Last Theorem, the only solution occurs when $x^2y^2z^2=0$ . Thus, no solution exists.

Bonus

What about $\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{1}{w^3}$ ? Does the same result still apply?

X X
May 6, 2018

$(yz)^3+(xz)^3=(xy)^3$ ,because there doesn't exist a triplet $(a,b,c)$ that satisfy $a^3+b^3=c^3$ ,so there is no $(x,y,z)$ exist