Why is it necessary that different observers agree on the value of the spacetime interval $ds^2$? What's the physical reason that all (inertial) observers agree on the value of the spacetime interval
$$ds^2 = (c dt)^2 - dx^2 - dy^2 -dz^2 \, ?$$
What would be the physical implications if different (inertial) observers wouldn't find different values of $ds^2$, analogous to how they find different time intervals and different distances?
EDIT: none of the linked questions (to which this question supposedly is a duplicate) explain why a constant speed of light implies that all observers agree on the value of the spacetime interval. 
 A: The physical implication would be that different inertial observers would find different values for the speed of light. It is the assumption that c is constant for all observers that leads to that particular equation for the interval.
A: Provided all observers agree when it is zero, they will all agree on the same speed of light.
As far as I can see, the universal speed of light per se imposes no other requirement.
But the universal speed of light is not the only input.  We also require there to be a single underlying reality behind the different observations made in the different frames.  Hence there must be a consistent transformation from one observation to another.  That is what the 'distance' formula provides.
A: FIRST POSTULATE OF SPECIAL RELATIVITY
The laws of physics are the same and can be stated in their simplest form in all inertial frames of reference.
SECOND POSTULATE OF SPECIAL RELATIVITY
The speed of light c is a constant, independent of the relative motion of the source.
Postulate number 1 is such, so that causality is presserved. And postulate number 2 is such, because that is the quest of the theory of relativity. To explore how a Universe with such qualities would look like.
Neither postulate can be proven or explored inside the theory cuz the theory is based on them, we will need to step out of relativity into some other theory to explore such questions. And thus far, no such theory exists.
A: space intervals in Newtonian mechanics
In Newtonian mechanics different observers can disagree on the position of events.  As an example, let's say I am $100$ m to the left of you.  An event, $A$, happens where I am.  I might give that event the coordinate $x_A=0$, and you might say that same event is at $x'_A = -100$ m.  A short time later a second event, $B$, happens halfway between us.  I say $x_B=+50$ m, and you say $x_B=-50$ m.
The thing to focus on is that $A$ and $B$ are real places in space. Everyone agrees that $A$ happened there, it's just we call there different things.  We can define $\vec{A}$ as a vector that denotes the location of the event.  Different observers express the same physical vector in their own coordinate systems, but they are all talking about the same physical location, the same geometrical object $\vec{A}$.
While we may disagree on the label of the locations of the two events, both you and I agree that $|\vec{B} - \vec{A}| = \Delta r = 50$ m.  The displacement vector $\Delta\vec{r}$ between $A$ and $B$ is a single geometrical object.  Everyone interacts with that distance, and in Newtonian physics everyone says $A$ and $B$ are the same distance apart.
So why is it that in Newtonian mechanics all observers agree on the value of the space interval?
$$\Delta\vec{r}\cdot\Delta\vec{r} = \Delta r^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$$
In Newtonian mechanics all observers have a shared physical reality that preserves distances.  Mathematically, distance is a scalar quantity, while position and displacement are vectors.  Different people can label the vectors differently in their own coordinates, but scalars don't have $x,y,z$ coordinates.  The length of a vector, which you can calculate from the dot-product of a vector with itself, is a scalar.
$$|d|^2 = \vec{d}\cdot\vec{d}$$
The length of position vectors is a bit trickier.  The length of the position I give to $A$ tells the distance between me and $A$.  You would agree on that distance, $0$ m, but that is not the same the length of your position vector, $100$ m, the distance between you and $A$.
spacetime intervals in special relativity
A similar thing happens in SR.  People disagree on the labels (coordinates) they give to events in spacetime.  But spacetime events are real places in spacetime.  All observers occupy the same shared physical reality.  Everyone agrees that $A$ and $B$ happened there and there, they just label those places differently.  The spacetime displacement between those two events is a single geometrical object,
$$\Delta\vec{s} = \vec{B} - \vec{A}.$$
The length of that vector is still a scalar defined by its dot-product with itself,
$$\Delta\vec{s}\cdot\Delta\vec{s} = \Delta s^2.$$
The spacetime of SR has a new 4D dot-product, defined by the Minkowski metric.  It takes into account both spacial and temporal displacement.
$$\vec{a}\cdot\vec{b} = -c^2 a_t b_t + a_x b_x + a_y b_y + a_z b_z$$
So the length of the spacetime distance between $A$ and $B$ is
$$\Delta s^2 = -c^2 \Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2$$
This is the spacetime interval.  $\Delta s^2$ is a scalar, with no coordinates of its own.
In SR time intervals ($\Delta t$) and spacial displacements ($\Delta r$) are observer dependent, but the combined spacetime interval is not.  The 4D dot-product defines what things are invariant.  And in SR the form of the dot-product depends on the constancy of the speed of light.
You could make up a new dot-product, based on a metric that is not the Minkowski metric, and the form of the invariant interval in your new universe would be different.  But that new quantity wouldn't be useful for SR.
