Exercise 5.3.3 (Pythagorean triples in arithmetic or geometric progression)

Proof. Let $(x,y,z)\in {\mathbb{Z}}^3$ be a Pythagorean triple, so that $x^2+y^2=z^2$.

(a)

if $x,y,z$ form an arithmetic progression, then $y=\frac{x+z}{2}$. Thus $(x,z)$ is solution of $$\begin{array}{lll}\hfill x^2+{\left(\frac{x+z}{2}\right)}^2=z^2.& & \hfill \\ \hfill & & \hfill \end{array}$$

Then

$$\begin{array}{llll}\hfill 0& =4x^2+{(x+z)}^2-4z^2& \hfill & \\ \hfill & =5x^2+2\mathit{xy}-3z^2& \hfill & \\ \hfill & =(5x-3z)(x+z).& \hfill & \end{array}$$

Therefore $x+z=0$ or $5x-3z=0$.

If $x+z=0$, then $y=0$, and $(x,y,z)=(-a,0,a)$ for some $a\in \mathbb{Z}$, where $(-a,0,a)$ are indeed Pythagorean triples.

If $5x-3z=0$, then $3\mid 5x$, where $3\wedge 5=1$, thus $3\mid x$. There is some $a\in \mathbb{Z}$ such that $x=3a$. Therefore $3z=15a$, $z=5a$, and $y=(x+z)∕2=4a$, so

$$(x,y,z)=(3a,4a,5a),$$

which are Pythagorean triples. The Pythagorean triples whose terms form an arithmetic progression are

$$(-a,0,a),(3a,4a,5a),a\in \mathbb{Z},$$

so are integers multiples of the primitive triples $(-1,0,1)$ or $(3,4,5)$.

(b)

If $x,y,z$ form an geometric progression, then $y^2=\mathit{xz}$. Thus $(x,z)$ is solution of $$x^2+\mathit{xz}-z^2=0.$$

If $z=0$, then $x=0$ and $y=0$, so $(x,y,z)=(0,0,0)$.

if $z\ne 0$, then

$$\begin{array}{lll}\hfill {\left(\frac{x}{z}\right)}^2+\frac{x}{z}-1.& & \hfill \text{(1)}\end{array}$$

But the roots of the polynomial $X^2+X-1$ are $(-1-\sqrt{5})∕2$ and $(-1-\sqrt{5})∕2$, which are irrational numbers. Therefore $x∕z$ cannot be solution of (1).

Apart the trivial solution $(0,0,0)$, there is no Pythagorean triple whose terms form a geometric progression.

(I let Byzantine doctors discuss on the fact that $(0,0,0)$ form a geometric progression or not.)