(copied from my answer to Minkowski Metric Signature [with some modifications])
Here's an argument essentially due to Bondi.
It is physically motivated by radar measurements.
First, an introduction to Bondi's k-calculus.
(This is based on a diagram from Bondi's "E=mc2: An Introduction to Relativity" (http://www.worldcat.org/title/emc2-an-introduction-to-relativity/oclc/156217827),
which accompanied Bondi's series of lectures "E=mc2: Thinking Relativity Through", a series of ten lectures on BBC TV running from Oct 5 to Dec 7, 1963. It had a typo that I corrected.)
Two inertial observers (Bondi will call) Alfred and Brian meet at event O.
Alfred performs a radar measurement to assign coordinates to event P on Brian's worldline.
After a time $T$ on Alfred's wristwatch, he sends a light signal to Brian.
Brian receives the signal at a time $kT$ on Brian's watch (event P), where $k$ is a proportionality constant (independent of $T$). [This $k$ turns out to be the Doppler factor].
When this light-signal is reflected by Brian's worldline (at event P),
the reflected signal back arrives at Alfred's worldline when Alfred's watch reads $k(kT)$,
where the same factor of $k$ is used because of the Principle of Relativity.
(We've also used that the speed of light is the same for these observers.)
[Side note: These two triangles, with two timelike legs and one lightlike leg, are similar in Minkowski spacetime.]
So, Alfred can assign a time-coordinate and a space-coordinate to the distant event P (displacements from event O):
$$\Delta t_{P}=(\mbox{half of the elapsed time})=\frac{t_{rec}+t_{send}}{2}=\frac{k^2T+T}{2}$$
$$\Delta x_{P}=(\mbox{half of the roundtrip distance})=c\frac{t_{rec}-t_{send}}{2}=c\frac{k^2T-T}{2}.$$
By division, one can get $\quad v_{BA}=\displaystyle\frac{\Delta x_P}{\Delta t_P}=\frac{k^2-1}{k^2+1}\quad$ (independent of $T$),
which can be solved for $k$ to get the Doppler formula.
Note that
by addition: $\quad \Delta t_{P}+(1/c)\Delta x_{P}=t_{rec}\quad$, and
by subtraction: $\Delta t_{P}-(1/c)\Delta x_{P}=t_{send}$.
Now consider two inertial observers making radar measurements, assigning coordinates to a distant event (call it Q).
Each observer sends a light-signal and waits for its echo to be received, noting his wristwatch reading at these two events on his worldline.
(Geometrically, we have the light-cone of Q intersecting the two inertial worldlines that met at event O.)
[Side note: Although not necessary, event Q could be on the worldline of a third observer (call her Carol). Then these radar measurements would involve $k_{CB}$ and $k_{CA}$, relating Carol and Brian and Carol and Alfred.
The "$k$" used above in the first part and in the part below could be called
$k_{BA}$ to relate Brian and Alfred.]
(The diagram is from Bondi's "Relativity and Common Sense".)
Their wristwatch readings are related by
$$t'_{send}=k t_{send}$$
$$\left( \Delta t_Q' - \frac{\Delta x_Q'}{c}\right) = k\left( \Delta t_Q - \frac{\Delta x_Q}{c} \right)$$
and
$$t'_{rec}=\frac{1}{k} t_{rec}$$
$$\left( \Delta t_Q' + \frac{\Delta x_Q'}{c}\right) = \frac{1}{k}\left( \Delta t_Q + \frac{\Delta x_Q}{c} \right)$$
By multiplication, we get the following equation, which is independent of $k$:
$$t'_{send}t'_{rec}=t_{send} t_{rec}\mbox{... in terms of radar coordinates}$$
which can be written in terms of rectangular components as
$$\left({\bf \mbox{invariant square interval}}\right)=\left( \Delta t_Q'^2 - \frac{\Delta x_Q'^2}{c^2}\right)
=\left( \Delta t_Q^2 - \frac{\Delta x_Q^2}{c^2}\right),$$
with its minus-sign in front of the spatial coordinate.
(Calling this "the invariant square interval" and not "minus the invariant square interval" is the
choice of sign convention.)
(Side note: By addition and subtraction, one gets the Lorentz transformations.)
The reason why this method works is that we are working in the eigenbasis of
the Lorentz Transformation, where the the lightlike directions are the eigenvectors
and the Doppler factor and its reciprocal are the eigenvalues.
This is based on a blog entry that I contributed here
https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/
UPDATE to address the question in the comment by @sleep.
visit my Desmos visualization robphy v8e spacetime diagrammer for relativity (v8e-2021) - t-UP
- The "unit circle" (the "curve of constant square-interval 1") has the form
$$t^2-Ey^2=(1)^2$$
in the Euclidean ($E=-1$),
Galilean ($E=0$),
and
Minkowskian ($E=+1$) case.
- "Perpendicular to a radius vector" (i.e. a "right angle" in that geometry) is defined using the tangent to the circle.
- "Hypotenuse of a right-triangle" is defined as the side opposite the right-angle.
The relative-slope (with respect to the vertical)
in these cases is $v=3/5$.
For $E=-1$, this triangle with adjacent side of 1 has a hypotenuse (here, 1.162) that is longer than the adjacent side. Indeed, $$(hyp)=\frac{(adj)}{\cos\phi}\geq (adj) \mbox{ since $\cos\phi \leq 1$, where $\tan\phi=v$}.$$
For $E=0$, this triangle with adjacent side of 1 has a hypotenuse (here, 1) that is equal to the adjacent side.
For $E=+1$, this triangle with adjacent side of 1 has a hypotenuse (here, 0.8) that is shorter than the adjacent side. Indeed, $$(hyp)=\frac{(adj)}{\cosh\theta}\leq (adj) \mbox{ since $\gamma=\cosh\theta \geq 1$, where $\tanh\theta=v$}.$$
If you want to see a Pythagorean triple in the Minkowski case,
you can use my "Relativity on Rotated Graph Paper" method.
You can construct other triples on the diagram of the elastic-collision at https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
