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Edit 2 For good measure . . .

6. Hilbert 90

Consider, as above, the equation $X^2 + 5Y^2 = 1$. It shows that the element $X + Y \sqrt{-5}$ has norm $1$;by Hilbert 90, we must have$$X + Y \sqrt{-5} = \frac{a+b\sqrt{-5}}{a-b\sqrt{-5}} = \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{a^2 + 5b^2}.$$Dehomogenizing via $X = \frac xz$ and $Y = \frac yz$ yields the sameprojective parametrization as above, and we end up with thefamiliar formulas.

7. Binary Quadratic FormsThe equation $x^2 + 5y^2 = z^2$ tells us that the form $Q_0(X,Y) = X^2 + 5Y^2$with fundamental discriminant $\Delta = -20$ represents a square; this implies that $Q_0$ lies in the principal genus (which is trivialsince $Q_0$ is the principal form), and that the representations of $z^2$ by $Q_0$ come from composing representations of $z$ by forms $Q_1$ with $Q_1^2 \sim Q_0$ with themselves.

There are only two forms with discriminant $\Delta$ whose square is equivalent to $Q_0$: the principal form $Q_0$ itself and the form$Q_1(X,Y) = 2X^2 + 2XY + 3Y^2$. Thus either$$z = Q_0(a,b) = a^2 + 5b^2 \quad \text{or} \quad z = Q_1(a,b) = 2a^2 + 2ab + 3b^2.$$The formulas for Gauss composition of forms immediately provide us withexpressions for $x$ and $y$ in terms of $a$ and $b$, but they can alsobe checked easily by hand. In the first case, we get$$x^2 + 5y^2 = (a^2 + 5b^2)^2 = (a^2 - 5b^2)^2 + 5(2ab)^2,$$and in the second case we can reduce the equations to this oneby observing that $2Q_1(a,b) = A^2 + 5b^2$ with $A = 2a+b$, which gives$$x^2 + 5y^2 = \frac14\Big(A^2 + 5b^2\Big)^2 = \Big(\frac{A^2 - 5b^2}2\Big)^2 + 5(Ab)^2.$$

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Suppose you have a diophantine problem whose solution is connected with the structure of the p-class group of a number field K. Then you have the following options:

1. Use ideal arithmetic in the maximal order OK
2. Replace OK by a suitable ring of S-integers with trivial p-class group
3. Replace K by the Hilbert class field, which (perhaps) has trivial p-class group.

Experience with descent on elliptic curves has shown me that ultimately, the equations you have to solve in methods 1 and 2 are the same; moreover, the approach using ideals is a lot less technical than using factorial domains in S-integers (the class group relations come back in through the larger rank of the group of S-units). I am certain that the route via the Hilbert class field is even more technical: again, the unit group in the class field will produce more difficulties than a trivial class group will eliminate.

Edit. As an example illustrating my point in a very simple example, let me solve the diophantine equation $x^2 + 5y^2 = z^2$ in several different ways. I will always assume that $\gcd(x,y) = 1$.

1. Elementary Number Theory

The basic idea is factoring: from $5y^2 = (z+x)(z-x)$. Since $d = \gcd(z-x,z+x) = \gcd(2z,z+x) \mid 2$ we have $d = 1$ or $d = 2$; moreover we clearly have $z-x > 0$. This gives $z+x = da^2$, $z-x = 5db^2$ or $z+x = 5da^2$, $z-x = db^2$. Solving for $x$ and $z$ yields $$x = \pm \frac d2 (a^2 - 5b^2), \quad y = dab.$$

2. Parametrization

Set $X = \frac xz$ and $Y = \frac yz$; then $X^2 + 5Y^2 = 1$. Take the line $Y = t(X+1)$ through the obvious point $(-1,0)$; the second point of intersection is given by $$X = \frac{1-5t^2}{1+5t^2}, \quad Y = \frac{2t}{1+5t^2}.$$ Dehomogenizing using $t = \frac ba$ and $X = \frac xz$ etc. gives the projective parametrization $$(x:y:z) = (a^2-5b^2:2ab:a^2+5b^2).$$ If $ab$ is odd, all coordinates are even, and we find $$x = \frac12(a^2 - 5b^2), \quad y = ab;$$ if $a$ or $b$ is even we get $$x = a^2 - 5b^2, \quad y = 2ab$$ as above.

3. Algebraic Number Theory

Consider the factorization $$(x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2$$ in the ring of integers of the number field $K = {\mathbb Q}(\sqrt{-5}\,)$. The class number of $K$ is $2$, and the ideal class is generated by the prime ideal ${\mathfrak p} = (2,1+\sqrt{-5}\,)$.

The ideal $(x + y\sqrt{-5}, x - y\sqrt{-5}\,)$ is either $(1)$ or ${\mathfrak p}$; thus $$(x + y\sqrt{-5}\,) = {\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) = {\mathfrak b}^2$$ in the first and $$(x + y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak b}^2$$ in the second case.

The second case is impossible since the left hand side as well as ${\mathfrak a}^2$ are principal, but ${\mathfrak p}$ is not. We could have seen this immediately since $x$ and $y$ cannot both be odd.

In the first case, assume first that ${\mathfrak a} = (a + b\sqrt{-5}\,)$ is principal. Since the only units in ${\mathcal O}_K$ are $\pm 1$, this gives $x + y \sqrt{-5} = \pm(a+b\sqrt{-5}\,)^2$ and hence $$x = \pm (a^2 - 5b^2), \quad y = \pm 2ab.$$ If ${\mathfrak a}$ is not principal, then ${\mathfrak p}{\mathfrak a} = (a+b\sqrt{-5}\,)$ is, and from $({\mathfrak p}{\mathfrak a})^2 = 2(x+y\sqrt{-5}\,)$ we similarly get $$x = \pm \frac12(a^2 - 5b^2), \quad y = \pm ab.$$

4. S-Integers

The ring $R = {\mathbb Z}[\sqrt{-5}\,]$ is not a UFD, but $S = R[\frac12]$ is; in fact, $S$ is even norm-Euclidean for the usual norm in $S$ (the norm is the same as in $R$ except that powers of $2$ are dropped). It is also easily seen that $S^\times = \langle -1, 2 \rangle$. From $$(\ref{E5}) x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2$$ and the fact that the factors on the left hand side are coprime we deduce that $x + y\sqrt{-5} = \varepsilon \alpha^2$ for some unit $\varepsilon \in S^\times$ and some $\alpha \in S$. Subsuming squares into $\alpha$ we may assume that $\varepsilon \in {\pm 1, \pm 2}$. Setting $\alpha = \frac{a + b\sqrt{-5}}{2^t}$, where we may assume that $a$ and $b$ are not both even, we get $$x + y \sqrt{-5} = \varepsilon \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{2^{2t}}.$$ It is easily seen that we must have $t = 0$ and $\varepsilon = \pm 1$ or $t = 1$ and $\varepsilon = \pm 2$; a simple calculation then yields the same formulas as above.

5. Hilbert Class Fields

The Hilbert class field of $K$ is given by $L = K(i)$. It is not too difficult to show that $L$ has class number $1$ (actually it is norm-Euclidean), and that its unit group is generated by $i = \sqrt{-1}$ and $\omega = \frac{1+\sqrt{5}}2$ (we only need to know that these units and their product are not squares). From $$(\ref{E5}) x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2$$ and the fact that the factors on the left hand side are coprime in ${\mathcal O}_K$ we deduce that $x + y \sqrt{-5} = \varepsilon \alpha^2$. Subsuming squares into $\alpha^2$ we may assume that $\varepsilon \in {1, i, \omega, i\omega }$. Applying the nontrivial automorphism of $L/K$ to $x + y \sqrt{-5} = \varepsilon \alpha^2$ we find $\varepsilon \alpha^2 = \varepsilon' {\alpha'}^2$. Since the ideal ${\mathfrak a} = (\alpha)$ is fixed and since $L/K$ is unramified, the ideal ${\mathfrak a}$ must be an ideal in ${\mathcal O}_K$. Thus either ${\mathfrak a} = (a+b\sqrt{-5}\,)$ is principal in $K$, or ${\mathfrak p} {\mathfrak a} = (a+b\sqrt{-5}\,)$ is; in the second case we observe that ${\mathfrak p} = (1+i)$ becomes principal in ${\mathcal O}_L$.

Thus either $$x + y \sqrt{-5} = (a+b\sqrt{-5}\,)^2 \quad \text{or} \quad x + y \sqrt{-5} = i \Big(\frac{a+b\sqrt{-5}}{1+i}\,\Big)^2,$$ giving us the same formulas as above.

Avoiding ideal arithmetic in $K$ and only using the fact that ${\mathcal O}_L$ is a UFD seems to complicate the proof even more.

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Edit. As an example illustrating my point in a very simple example, let me solve the diophantine equation $x^2 + 5y^2 = z^2$ in several different ways. I will always assume that $\gcd(x,y) = 1$.

1. Elementary Number Theory

The basic idea is factoring: from $5y^2 = (z+x)(z-x)$. Since $d = \gcd(z-x,z+x) = \gcd(2z,z+x) \mid 2$ we have $d = 1$ or $d = 2$; moreover we clearly have $z-x > 0$. This gives $z+x = da^2$, $z-x = 5db^2$ or $z+x = 5da^2$, $z-x = db^2$. Solving for $x$ and $z$ yields$$x = \pm \frac d2 (a^2 - 5b^2), \quad y = dab.$$

2. Parametrization

Set $X = \frac xz$ and $Y = \frac yz$; then $X^2 + 5Y^2 = 1$. Take the line $Y = t(X+1)$ through the obvious point $(-1,0)$; the second point of intersection is given by $$X = \frac{1-5t^2}{1+5t^2}, \quad Y = \frac{2t}{1+5t^2}.$$Dehomogenizing using $t = \frac ba$ and $X = \frac xz$ etc. givesthe projective parametrization$$(x:y:z) = (a^2-5b^2:2ab:a^2+5b^2).$$If $ab$ is odd, all coordinates are even, and we find$$x = \frac12(a^2 - 5b^2), \quad y = ab;$$if $a$ or $b$ is even we get$$x = a^2 - 5b^2, \quad y = 2ab$$as above.

3. Algebraic Number Theory

Consider the factorization $$(x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2$$in the ring of integers of the number field $K = {\mathbb Q}(\sqrt{-5}\,)$.The class number of $K$ is $2$, and the ideal class is generated bythe prime ideal ${\mathfrak p} = (2,1+\sqrt{-5}\,)$.

The ideal $(x + y\sqrt{-5}, x - y\sqrt{-5}\,)$ is either $(1)$ or ${\mathfrak p}$; thus $$(x + y\sqrt{-5}\,) = {\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) = {\mathfrak b}^2$$in the first and $$(x + y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak b}^2$$in the second case.

The second case is impossible since the left hand side as well as ${\mathfrak a}^2$ are principal, but ${\mathfrak p}$ is not. We could have seen this immediately since $x$ and $y$ cannot both be odd.

In the first case, assume first that ${\mathfrak a} = (a + b\sqrt{-5}\,)$ is principal. Since the only units in ${\mathcal O}_K$ are $\pm 1$, this gives $x + y \sqrt{-5} = \pm(a+b\sqrt{-5}\,)^2$ and hence$$x = \pm (a^2 - 5b^2), \quad y = \pm 2ab.$$If ${\mathfrak a}$ is not principal, then ${\mathfrak p}{\mathfrak a} = (a+b\sqrt{-5}\,)$ is,and from $({\mathfrak p}{\mathfrak a})^2 = 2(x+y\sqrt{-5}\,)$ we $$x = \pm \frac12(a^2 - 5b^2), \quad y = \pm ab.$$

4. S-Integers

The ring $R = {\mathbb Z}[\sqrt{-5}\,]$ is not a UFD, but $S = R[\frac12]$ is;in fact, $S$ is even norm-Euclidean for the usual norm in $S$(the norm is the same as in $R$ except that powers of $2$ are dropped).It is also easily seen that $S^\times = \langle -1, 2 \rangle$. From(\ref{E5}) and the fact that the factors on the left hand side are coprime we deduce that $x + y\sqrt{-5} = \varepsilon \alpha^2$ for some unit $\varepsilon \in S^\times$ and some $\alpha \in S$. Subsuming squares into$\alpha$ we may assume that $\varepsilon \in {\pm 1, \pm 2}$. Setting$\alpha = \frac{a + b\sqrt{-5}}{2^t}$, where we may assume that $a$ and $b$ are not both even, we get$$x + y \sqrt{-5} = \varepsilon \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{2^{2t}}.$$It is easily seen that we must have $t = 0$ and $\varepsilon = \pm 1$ or $t = 1$ and $\varepsilon = \pm 2$; a simple calculation then yields the same formulas as above.

5. Hilbert Class Fields

The Hilbert class field of $K$ is given by $L = K(i)$. It is nottoo difficult to show that $L$ has class number $1$ (actually it isnorm-Euclidean), and that its unit group is generated by $i = \sqrt{-1}$and $\omega = \frac{1+\sqrt{5}}2$ (we only need to know that theseunits and their product are not squares). From (\ref{E5}) and the factthat the factors on the left hand side are coprime in ${\mathcal O}_K$ we deduce that $x + y \sqrt{-5} = \varepsilon \alpha^2$. Subsuming squares into $\alpha^2$ we may assume that $\varepsilon \in { i, \omega, i\omega }$. Applying the nontrivial automorphism of $L/K$ to $x + y \sqrt{-5} = \varepsilon \alpha^2$ we find$\varepsilon \alpha^2 = \varepsilon' {\alpha'}^2$. Since the ideal ${\mathfrak a} = (\alpha)$ is fixed and since $L/K$ is unramified, the ideal ${\mathfrak a}$ must be an ideal in ${\mathcal O}_K$. Thus either ${\mathfrak a} = (a+b\sqrt{-5}\,)$ is principal in $K$,or ${\mathfrak p} {\mathfrak a} = (a+b\sqrt{-5}\,)$ is; in the second case we observethat ${\mathfrak p} = (1+i)$ becomes principal in ${\mathcal O}_L$.

Thus either $$x + y \sqrt{-5} = (a+b\sqrt{-5}\,)^2 \quad \text{or} \quad x + y \sqrt{-5} = i \Big(\frac{a+b\sqrt{-5}}{1+i}\,\Big)^2,$$giving us the same formulas as above.

Avoiding ideal arithmetic in $K$ and only using the fact that ${\mathcal O}_L$ is a UFD seems to complicate the proof even more.

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