= Solution
For nonzero <ideals>, <unique factorization of ideals in a number field> converts $AB=C^3$ to $v_P(A)+v_P(B)=3v_P(C)$. Coprimality says at most one of the first two exponents is positive, so each is divisible by three. Dividing its exponents by three constructs <ideals> whose cubes are $A,B$. If a zero <ideal> is allowed, coprimality forces the other <ideal> to be the whole <ring> and the conclusion is immediate with the zero <ideal> as its own cube.
To compute the <Ideal class group of Q of square root minus thirty>, put $\theta=\sqrt{-30}$. The <ring of integers> is $\mathbb Z[\theta]$, with <discriminant> $-120$. The given <Minkowski bound for ideal classes> is $2\sqrt{120}/\pi=4\sqrt{30}/\pi<7$. Hence the <ideal class group> is generated by <prime ideals> with <ideal norm> below seven. The rational primes two, three and five all ramify, giving unique <prime ideals> $P_2=(2,\theta)$, $P_3=(3,\theta)$, $P_5=(5,\theta)$ with $P_i^2=(i)$ and <ideal norm> $i$. There is no additional <prime ideal> of <ideal norm> four because two ramifies with <residue degree> one; six is not a <prime power>. Thus these three classes generate and each has order at most two.
As $\theta^2=-30$, <unique factorization of ideals in a number field> gives $(\theta)=P_2P_3P_5$, so their three classes multiply to one. None is principal: a generator of one would be an integral element $a+b\theta$ of <field norm> $a^2+30b^2$ equal to two, three or five, and none of those <field norm> equations has <integer> solutions. The three nontrivial classes are therefore distinct, with the third the product of the first two. Hence
$$
\boxed{\operatorname{Cl}(\mathbb Q(\sqrt{-30}))\cong C_2\times C_2.}
$$
To rule out <integer> solutions, suppose $x^2+30=y^3$. Necessarily $y>0$. If a prime $p\mid30$ divided $y$, then reduction modulo $p$ would force $p\mid x$. Because thirty is squarefree, the valuation of $x^2+30$ at that prime would be exactly one, while that of $y^3$ is a positive multiple of three, a contradiction. Thus $\gcd(y,30)=1$.
If a <prime ideal> divided both $(x+\theta)$ and $(x-\theta)$, it would divide their difference $(2\theta)$ and their product $(y)^3$. Its underlying rational prime would divide $N(2\theta)=120$, hence divide thirty, and would also divide $y$. This contradicts the gcd just proved, so the two <ideals> are coprime. Their product is $(y)^3$, so the first part gives $(x+\theta)=A^3$ for an <ideal> $A$.
In a <ideal class group> of exponent two, $[A]^3=1$ forces $[A]=1$. Thus $A=(a+b\theta)$ with $a,b\in\mathbb Z$. Equality of principal <ideals> gives $x+\theta=\varepsilon(a+b\theta)^3$ for a <unit> $\varepsilon$. The only units are $\pm1$, since $a^2+30b^2=1$ forces $b=0$, $a=\pm1$; both signs are cubes and can be absorbed into the generator. Comparing the coefficient of $\theta$ after cubing now gives
$$
1=3a^2b-30b^3=3b(a^2-10b^2),
$$
which is impossible modulo three. Therefore
$$
\boxed{x^2+30=y^3\text{ has no solutions in }\mathbb Z^2.}
$$
Back to article page
