Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 20F (Number Fields)

Put $s=\sqrt{-17}$. The ring of integers of a number field is $\mathbb{Z}[s]$ and the field discriminant is $-68$. The Minkowski bound for ideal classes says that each ideal class has an integral ideal representative of norm at most $(2/\pi )\sqrt{68}<6$.

The prime two ramifies as $(2)={\mathfrak{p}}_2^2$, where ${\mathfrak{p}}_2=(2,1+s)$. Three splits as $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, choosing ${\mathfrak{p}}_3=(3,1+s)$, whereas five is inert because $-17\equiv 3(\mathrm{mod}5)$ is not a square. The norm bound therefore leaves only the classes of $1$, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\overline{\mathfrak{p}}}_3$; a norm-four ideal is ${\mathfrak{p}}_2^2$ and is principal.

The ideal ${\mathfrak{p}}_2$ is not principal, since an element of norm two would solve $a^2+17b^2=2$. On the other hand $N(1+s)=18$, and $1+s$ is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$. Unique prime ideal factorization and the norm give $(1+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Thus $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2]$, which has order two, and

For the Diophantine equation, $x$ must be positive. If $x$ were even, $y^2\equiv 15(\mathrm{mod}32)$, impossible for a square. Thus $x$ is odd and $y$ even. If $17\mid x$, then $17\mid y$, but $y^2+17$ would have 17-adic valuation one while $x^5$ has valuation divisible by five, also impossible.

The ideals $(y+s)$ and $(y-s)$ are consequently coprime. A common prime divides $2s$ and hence lies over two or seventeen, both excluded above. Their product is $(x{)}^5$, so unique prime ideal factorization gives $(y+s)={\mathfrak{a}}^5$. Since the ideal class group has order four, $[\mathfrak{a}{]}^5=1$ forces $[\mathfrak{a}]=1$. Write $\mathfrak{a}=(a+bs)$ with integers $a,b$. The only units are $\pm 1$ by their norm, and the sign can be absorbed into the fifth power. Hence $y+s=(a+bs{)}^5$.

Comparing the coefficient of $s$ givesFor $b=1$, the bracket would be one, but it is four modulo five. For $b=-1$, the bracket would be minus one, but it is one modulo three for either possible square residue of $a$. Both cases are impossible. Therefore there are no integer solutions.