Cubic Diophantine Family

The left side of this equation looks very similar to the well-known identity $\begin{aligned} a^3+b^3+c^3-3abc &= (a+b+c)(a^2+b^2+c^2-ab-bc-ac) \\ &= (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) \end{aligned}$ where $\omega$ is a complex primitive cube root of unity, and substituting $a = x, b = y \sqrt[3]{2}, c = z\sqrt[3]{4}$ turns the equation into $(x+y\sqrt[3]{2}+z\sqrt[3]{4})(x+y\omega\sqrt[3]{2}+z\omega^2\sqrt[3]{4})(x+y\omega^2\sqrt[3]{2}+z\omega\sqrt[3]{4}) = 1$ This is recognizable as the product of the conjugates of the element $\alpha = x+y\sqrt[3]{2} + z\sqrt[3]{4}$ , also known as the norm of $\alpha$ . So this equation just says $N(\alpha) = 1$ .

The point is that norm is multiplicative: $N(\alpha\beta) = N(\alpha)N(\beta)$ . (I suppose this is not exactly elementary, but it reduces to a certain extremely tedious algebraic identity that could be proved directly.)

Now note that $(1,1,1)$ is a solution, i.e. $\alpha = 1+\sqrt[3]{2} + \sqrt[3]{4}$ has norm $1$ . Then so does any power of $\alpha$ . Then $\alpha^6 = 1081 + 858\sqrt[3]{2} + 681 \sqrt[3]{4}$ gives the $x$ -value we are looking for. So the answer is $\fbox{1081}$ .

Notes:

(1) It is in fact also true that every solution comes from a power of $\alpha$ , but this requires some deep algebraic number theory.

(2) Another way to say all this is: if $(x,y,z)$ is a solution to the original equation, then so is $(x+2y+2z,x+y+2z,x+y+z)$ . (And the inverse of this linear transformation: $(2z-x,x-y,y-z)$ .) This can be checked completely elementarily, but the motivation is that this corresponds to multiplying $x+y \sqrt[3]{2} + z\sqrt[3]{4}$ by the "fundamental unit" $1+\sqrt[3]{2} + \sqrt[3]{4}$ . You can repeatedly apply this linear transformation in Excel to compute $\alpha^6$ quite quickly.

This problem can be solved with some pretty fancy number theory: It turns out that the integer solutions $x,y,z$ satisfy an equation $(1+\sqrt[3]{2}+\sqrt[3]{4})^n=x+y\sqrt[3]{2}+z\sqrt[3]{4}$ for some integer $n$ . For $n=6$ we find the solution $(x,y,z)=(1081,858,681)$ , the only solution with $x$ on the required interval. (We have $x=281$ for $n=5$ and $x=4159$ for $n=7$ .)