# Solving the homogeneous Diophantine equation $x^3 + 2y^3 = 7z^3$ for $x,y,z \in \mathbb{Q}$

I want to solve the homogeneous Diophantine equation $$x^3 + 2y^3 = 7z^3$$ for $$x,y,z \in \mathbb{Q}$$.

First note that $$(x,y,z) = (0,0,0)$$ is a solution.

For further solutions it suffices to search for solutions in $$\mathbb{Z}^3$$, because if we have a solution in $$\mathbb{Q}^3$$ we can always multiply by the product of the denominators to find a solution in $$\mathbb{Z}^3$$.

Note that third powers are special in $$\mathbb{Z}/7\mathbb{Z}$$, since $$3 |\phi(7)=6$$. And we have that $$x^3 \in \{0,\pm1\}$$ for $$x \in \mathbb{Z}/7\mathbb{Z}$$. So we reduce the equation $$\mod{7}$$. To find $$x^3 \equiv y^3 \equiv 0 \mod{7}$$. Now let $$x^3 = a\, 7$$ and $$y = b \, 7$$ for some $$a,b \in \mathbb{Z}$$. We substitute this back in the original equation to find $$a\,7 + 2\, b\,7=7z^3$$, or $$a+2b=z^3$$.

At this point I'm stuck. I don't know whether it was a good idea to substitute the $$a$$ and $$b$$, because we now seem to lose some information.

It might be an idea to look modulo other primes, but I don't know which, as only $$7$$ seemed to make sense.

It is better to write $$x=7a$$ and $$y=7b$$ (since $$7\mid x^3$$ we have $$7\mid x$$ and the same for $$y$$), so we get $$7^3a^3+2\cdot 7^3b^3 = 7z^3$$ so $$7\mid z$$ and thus $$z=7c$$ for some integer $$z$$, so we get basicly the same equation as before:

$$a^3+2b^3=7c^3$$

now you can procede this inifitly times. But if $$x,y,z>0$$ (or $$a,b,c$$) this is impossible. So $$a=b=c=0$$.

• I don't get what you're doing in the first equation; don't you mean $7^3a^3+2\cdot 7^3b^3 = 7z^3$? – Algebear Dec 16 '18 at 23:58
• Yes, thank you. @GuusPalmer – Maria Mazur Dec 17 '18 at 8:50
• Btw, couldn't you use this same principle to prove Fermat's last theorem for $n=3$? – Algebear Dec 17 '18 at 10:44

It seems to me that your problem can be much generalized, using more powerful tools from ANT such as Eisenstein's cubic recipocity law. Consider the diophantine equation (1) $$x^3+qy^3=pz^3$$, where $$p,q$$ are two distinct given primes, et let us solve it in two steps:

1) Suppose that we can show that (1) implies that $$x\equiv y\equiv 0$$ mod $$p$$. Then we can do infinite descent exactly as in the answer of @greedoid to conclude that (1) has no non trivial solution.

2) To go on, let us assume $$y \neq 0$$ mod $$p$$. Then the reduction of (1) modulo $$p$$ is equivalent to $$q\equiv t^3$$ mod $$p$$, $$t\in \mathbf Z$$, and we must distinguish three cases: (i) If $$p=3$$, Fermat's little theorem says that our congruence always has a solution ; (ii) If $$p\equiv -1$$ mod $$3$$, taking cubic powers induces an automorphism of $$(\mathbf Z/p\mathbf Z)^*$$, so our congruence is again solvable ; (iii) If $$p\equiv 1$$ mod $$3$$, the arithmetic of the Eisenstein ring $$R=\mathbf Z[\omega]$$, where $$\omega$$ is a primitive 3-rd root of unity, enters the game. It is classically known that $$R$$ is a PID (and even an euclidian ring), and that $$p$$ splits as a product $$p=\pi \bar\pi$$ , where $$\pi$$ is a prime of $$R$$. Then our congruence is solvable iff $$(\frac q\pi)_3=1$$, where $$(\frac ..)_3$$ denotes the cubic Legendre symbol. For the definition and properties of this symbol, a convenient reference is David Cox's book " Primes of the form $$x^2+ny^2$$ ", chap. 1, §4. Supposing further that $$q\equiv \pm 1$$ mod $$3$$, the cubic reciprocity law asserts that $$(\frac q\pi)_3(\frac \pi q )_3=1$$, so that our congruence is solvable iff $$(\frac \pi q )_3=1$$. By formula (4.10) in loc. cit., this is equivalent to $$(\frac \pi q )_3 \equiv \pi ^{(q^2-1)/3}$$mod $$q$$, and finally our congruence (coming from $$y\neq 0$$ mod $$p$$) is solvable iff $$p\equiv 1$$ mod $$3$$ and $$\pi ^{(q^2-1)/3} \equiv 1$$ mod $$q$$.

Summarizing, the infinite descent in 1) can be performed iff $$p\equiv 1$$ mod $$3$$ and $$\pi ^{(q^2-1)/3} \neq 1$$ mod $$q$$. You can easily check that this criterion contains your special case $$p=7, q=2$$.

NB: The extra hypothesis $$q\equiv \pm 1$$ mod $$3$$ is not necessary, but without it the cubic reciprocity law becomes more complicated to use.