= Solution
Work in $\mathcal O_K=\mathbb Z[\sqrt{-d}]$, since $d\equiv2\pmod4$. First $x$ must be odd: if it were even, coprimality would make $a$ odd, so $x^2+a^2d\equiv2\pmod4$, which cannot be an odd power $y^p$ for $p\ge3$. Thus $y$ is odd too. Also no odd prime $\ell$ can divide both $x$ and $d$. Such a prime does not divide $a$, and square-freeness of $d$ gives $v_\ell(x^2+a^2d)=1$, whereas $v_\ell(y^p)$ is divisible by $p$. Hence $\gcd(x,d)=1$.
Put $\alpha=x+a\sqrt{-d}$. Suppose a <prime ideal> $P$ divides both $(\alpha)$ and $(\overline\alpha)$, and let $\ell$ be the rational prime below it. It then contains $2x$ and $2a\sqrt{-d}$. The case $\ell=2$ is impossible because $N((\alpha))=y^p$ is odd. If $\ell$ is odd, the first inclusion gives $\ell\mid x$, and $\gcd(a,x)=1$ makes $a$ nonzero in the residue <field>. The second inclusion therefore gives $\sqrt{-d}\in P$, so $\ell\mid d$, contradicting the previous paragraph. Thus $(\alpha)$ and $(\overline\alpha)$ are coprime <ideals>. In particular they generate the <unit ideal>, the precise meaning of coprimality here; no unique factorization of elements is assumed.
Their product is $(y)^p$. Unique <prime ideal factorization> and disjoint prime support imply $(\alpha)=A^p$ for an <integral ideal> $A$. In the <ideal class group>, $[A]^p=1$. Since $p$ does not divide the <class number>, <Lagrange's theorem> forces $[A]=1$. Write $A=(u+v\sqrt{-d})$. Equality of the principal <ideals> gives $\alpha=\varepsilon(u+v\sqrt{-d})^p$. The only units are $\pm1$, since a unit satisfies $u^2+dv^2=1$ and $d\ge2$. Both signs are $p$th powers for odd $p$, so they can be absorbed into the generator:
$$
\boxed{x+a\sqrt{-d}=(u+v\sqrt{-d})^p\quad(u,v\in\mathbb Z).}
$$
Apply the <class-number obstruction from a ramified odd exponent>. For the equation with $a=1$ and $p\mid d$, compare the coefficients of $\sqrt{-d}$ in this identity. The coefficient on the right is
$$
v\sum_{\substack{1\le j\le p\\j\text{ odd}}}
\binom pj u^{p-j}v^{j-1}(-d)^{(j-1)/2}.
$$
Its $j=1$ term is divisible by $p$, and every $j\ge3$ term contains a factor $d$, so the whole coefficient is divisible by $p$. The coefficient on the left is one, a contradiction. Thus \b[there are no positive <integer> solutions under the stated class-number and divisibility assumptions].
Finally,
$$
\boxed{13^2+174=7^3.}
$$
The <integer> $174=2\cdot3\cdot29$ is square-free, is two modulo four, and is divisible by three. If its imaginary quadratic <class number> were not divisible by three, the preceding result with $p=3$ would prohibit this solution. Consequently
$$
\boxed{3\mid h(\mathbb Q(\sqrt{-174})).}
$$
Back to article page
