Definition of spacetime interval We know that a spacetime vector $x:= (\vec{x}, ct)$, where $c$ is the speed of light. Why is the interval $I$ in spacetime defined as
$$ I=-(\Delta t)^2 + \frac{1}{c^2}\left[ (\Delta x)^2+(\Delta y)^2+(\Delta z)^2 \right] ?$$
More concretely,
(1) Why is $I$ defined in terms of squared components (without the square root)? If it were $I^2$ then it'd be more clear.
(2) Why does division by $c^2$ happen in $I$? What is $c^2 I$ then?
Would appreciate some insights.
 A: 
I have written the answer with the following expression in mind as the expression for interval: $\Delta t^2 - \dfrac{1}{c^2}(\Delta x^2 + \Delta y^2 + \Delta z^2)$. In my experience, this is a rather common form of interval than the one stated by the OP. Whenever I write something like "the form you mentioned", it really means to refer to the above-stated expression. 

The stated quantity is a Lorentz invariant. This quantity, calculated for a pair of events, remains the same no matter which inertial observer measures it. Thus, it is a good idea to think of this quantity as a property of the pair of events itself, over and above its measurement by any observer. This is the essential reason to define this quantity as the interval - an analog of the Euclidean distance in Minkowskian spacetime. But, as you have partially pointed out, any bijective mapping of the stated quantity would satisfy all the above properties. So, why not define any one of them as the interval? To sum it up in one sentence: There is not much Physics in choosing this particular expression but it is rather a matter of convention. In fact, some variations of this quantity are also equally accepted as the interval. The most famous is the negative of this quantity (up to the factor of $c^2)$. Still, the origins of all the conventions are not stupid and there are good explanations for the particular questions you have asked: 
(1) It is used in the squared form because the quantity can be negative as well as positive. Thus, taking the square-root would introduce unnecessary work with imaginary quantities popping up everywhere. Another reason is that in Tensorial formulation of Special Relativity, which is crucial for its extension to General Relativity, this form finds a very elegant and easily manipulatable expression. Taking its root would again introduce unnecessary complications in the Tensorial representation.
(2) The stated expression is helpful to find out the proper time between two timelike events. To give this quantity temporal dimension, the division by $c^2$ is done. When you use the negative of the above formula, it is meant to be used to find out spacetime interval between two spacelike events. In that case, it is appropriate to give it spatial dimension and thus, indeed, the $c^2$ is multiplied with $t^2$ instead of dividing $x^2+y^2+z^2$ by $c^2$. All this mess is thrown out in any serious theoretical work by introducing the natural units where $c=1$. 
A: 
Why is $I$ defined in terms of squared components (without the square root)? If it were $I^2$ then it'd be more clear.

Look at Pythagoras's theorem,
$${\displaystyle a^{2}+b^{2}=c^{2}},$$ 
As Rudyard says, you can consider $I$  squared as a (modified) 4 dimension version of the 3 dimension expression above.

Why does division by $c^2$ happen in $I$? What is $c^2I$ then?

Relativity comes with the postulate that there is no absolute rate of the passage of time, so one may write:
$${\displaystyle ds^{2}=c^{2}d\tau ^{2}-dx_{\tau }^{2}-dy_{\tau }^{2}-dz_{\tau }^{2}},$$
$$ {\displaystyle d\tau ={\frac {ds}{c}}} $$
This is the way relativity deals with the time measured by the moving observer, also known as proper time $\tau$, analogous to time = distance/velocity on Earth.
If you read Wikipedia Proper Time or Hyperphysics you will get a more detailed explanation and derivation of the above expressions.
