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:


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

  • 1
    $\begingroup$ I don't get what you're doing in the first equation; don't you mean $7^3a^3+2\cdot 7^3b^3 = 7z^3$? $\endgroup$ – Algebear Dec 16 '18 at 23:58
  • $\begingroup$ Yes, thank you. @GuusPalmer $\endgroup$ – Maria Mazur Dec 17 '18 at 8:50
  • $\begingroup$ Btw, couldn't you use this same principle to prove Fermat's last theorem for $n=3$? $\endgroup$ – 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.


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