Past exam of the number theory course of the University of Oxford / 2010 / B / 6 / Solution

Equality $(\alpha )=(\beta )$ gives $\alpha =u\beta$ and $\beta =v\alpha$ for some $u,v\in {\mathcal{O}}_K$. Substitution and cancellation of the nonzero $\alpha$ yield $uv=1$. Thus $u$ is a unit and $\alpha =u\beta$.

For the ideal-power assertion, suppose first that all ideals are nonzero. A prime ideal divides at most one $Q_i$ because the $Q_i$ are pairwise coprime. Taking its valuation in the given equality givesSince $\gcd (n_i,k)=1$, every exponent $v_P(Q_i)$ is divisible by $k$. Dividing all its prime-factorization exponents by $k$ constructs an integral ideal $I_i$ with $Q_i=I_i^k$. If a $Q_i$ is zero, pairwise coprimality forces all other $Q_j$ to be the unit ideal; taking $I_i=0$ and the other $I_j={\mathcal{O}}_K$ covers that degenerate case too.

For the first Diophantine equation, $x$ cannot be odd: then $x^2+17\equiv 2(\mathrm{mod}8)$, while a cube modulo eight is $0$ or odd. Hence $x$ is even and $y$ is odd. Also $17\nmid x$, since otherwise $v_{17}(x^2+17)=1$, impossible for a cube.

In $\mathcal{O}=\mathbb{Z}[s]$, $s=\sqrt{-17}$, factorThe two principal ideals are coprime. A common prime ideal would divide their sum $2x$ and difference $2s$, and hence lie above $2$ or $17$; odd norm excludes $2$, while $17\nmid x$ excludes $17$. Their coprime product is the cube $(y{)}^3$, so the ideal-power assertion makes $(x+s)=I^3$. The class group has order $4$, while $[I{]}^3=1$, so $[I]=1$. Write $I=(a+bs)$ for integers $a,b$.

The only units are $\pm 1$, since their norm equation is $a^2+17b^2=1$. Equality of principal ideals therefore gives $x+s=\pm (a+bs{)}^3$. Absorb the sign into $a+bs$. Comparing the coefficient of $s$ yieldsThus $b=1$ or $-1$. The first gives $a^2=6$ and the second gives $3a^2=16$, both impossible for an integer $a$. Consequently the first equation has no integral solutions.

For the product equation, odd $x$ gives $(x^2+17)(9x^2+17)\equiv 4(\mathrm{mod}8)$, again not a cube. If $17\mid x$, each factor has $17$-adic valuation one, so the product has valuation two, also impossible. Thus any hypothetical solution has $x$ even and $17\nmid x$. The two positive factors are odd, and their greatest common divisor dividesIt must therefore be one. A product of coprime positive integers is a cube only if each factor is a cube. This would make $x^2+17$ a cube, contradicting the result already proved. Hence the product equation also has no integral solutions.