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Put $\omega =(1+\sqrt{229})/2$. Since $229\equiv 1(\mathrm{mod}4)$ is squarefree,We first compute the ideal class group, rather than assume the proposed extension has the full Hilbert-class-field degree.

The real quadratic Minkowski bound for ideal classes is $\sqrt{229}/2<8$. Thus every ideal class has an integral representative of norm at most seven. The prime two is inert because $229\equiv 5(\mathrm{mod}8)$, while seven is inert because $229\equiv 5(\mathrm{mod}7)$ is a nonsquare. The inert ideal $(2)$ has norm four and is principal; a prime above seven has norm forty-nine. The only possible nonprincipal prime factors of these small representatives are therefore primes above three and five.

Reduction of $X^2-X-57$ gives roots $0,1$ modulo three and $2,4$ modulo five. WriteTheir conjugate pairs multiply to $(3)$ and $(5)$. Direct norm calculations giveThe element $6+\omega$ vanishes at the residue roots $0$ modulo three and $4$ modulo five, and not at the conjugate roots. Its absolute norm forces each exponent to be one. Likewise $15+2\omega$ vanishes only at the root zero modulo three, and its norm forces exponent three. ThusThese also give the factorizations suggested by the hint: $(13+\sqrt{229})=(2)PQ$ and $(16+\sqrt{229})=P^3$. The classes of $P,Q$ generate the ideal class group; the first relation makes $[Q]=[P{]}^{-1}$ and the second makes $[P{]}^3=1$. Hence the class number divides three.

It is not one. The algebraic integerso it is a unit. If $P$ were principal, its generator $a=x+y\omega$ would have norm $\pm 3$. Multiply $a$ by a suitable integer power of $ϵ$ so thatIts conjugate then also has absolute value at most $\sqrt{3ϵ}$ because $|a{a}^{\prime }|=3$. ThereforeThe final inequality is $12ϵ<229$, immediate from the displayed value of $ϵ$. Thus the integer $y$ is zero, which makes $|N(a)|=x^2=3$ impossible. ConsequentlyThis proves the ideal class group of Q of square root 229. No claim that $ϵ$ is a fundamental unit is needed. Its norm minus one also realizes all real sign patterns after multiplication by $-1$, so the ordinary and narrow class numbers agree.

Now let $f(X)=X^3-4X-1$ and let $H$ be its splitting field. The possible rational roots $\pm 1$ are not roots, so $f$ is irreducible. Its discriminant isThus its Galois group is $S_3$, since an irreducible cubic has group $A_3$ precisely when its discriminant is a square. Its quadratic subfield is $\mathbb{Q}(\sqrt{229})=K$, and $H/K$ is cyclic of degree three. If $\alpha$ is one root, then $H=K(\alpha )$.

For every rational prime $\ell \ne 229$, the reduction of $f$ is separable. Its roots lift in an unramified extension containing its residue splitting field, by Hensel's lemma. Hence its splitting field is unramified over $\mathbb{Q}$ at all these primes, and consequently over $K$ there.

At 229 there is exactly one double residue root and one simple root:The simple root lifts to $r\in {\mathbb{Q}}_{229}$, and division gives $f=(X-r)q$ with $q$ monic quadratic. The discriminant-product identity isHere $q(r)=f^{\prime }(r)$ is a unit, since $r$ has the simple residue root. Therefore the quadratic discriminant has valuation one, and its root field is the ramified quadratic field ${\mathbb{Q}}_{229}(\sqrt{229})$. This is already the completion of $K$. The local splitting field has no further extension over it: the simple root lies in ${\mathbb{Q}}_{229}$ and the other two lie in that quadratic field. Thus $H/K$ is unramified even at 229. Equivalently its rational inertia is a transposition and has trivial intersection with $A_3$.

The positive cubic discriminant gives three distinct real roots, so $H$ is totally real and there is no infinite ramification over $K$. We have exhibited an everywhere-unramified cyclic extension of degree $h_K=3$. The Hilbert-class-field characterization in part (i) now proves

Finally let $p\ne 229$ be rational prime. A prime ideal of norm $p$ is principal if and only if its Artin symbol in $H_K/K$ is trivial. If it is principal, a generator has norm $\pm p$; multiplying by the norm-minus-one unit if necessary gives norm $+p$. Conversely an element of norm $p$ generates a prime ideal of norm $p$. Hence representation by $x^2+xy-57y^2$ is equivalent to $p$ splitting in $K$ with trivial Artin symbol in $H_K/K$, or to complete splitting of $p$ in $H_K/\mathbb{Q}$.

Let ${\tau }_p\in S_3$ be its rational Frobenius, defined up to conjugacy. Its restriction to the discriminant quadratic field is its sign. The splitting criterion for that quadratic field isFor odd $p$ this follows by quadratic reciprocity from the character of discriminant 229; for $p=2$ both criteria are negative since $229\equiv 5(\mathrm{mod}8)$. The polynomial is separable modulo $p$, and the degrees of its factors are the cycle lengths of ${\tau }_p$. Thus a root modulo $p$ means that ${\tau }_p$ fixes a root. An even permutation of three letters with a fixed point is the identity: a nonidentity three-cycle has none. Therefore the two stated conditions together are exactly complete splitting in $H_K$, andThis is prime representation by the norm form of Q of square root 229, with the norm sign and the prime two explicitly covered.