Is it true that the difference of two matrices is equal to a squared (multiplied by itself) matrice subtracted from another squared matrice? Why?

I'm not sure exactly what you are asking, but let's take a look...

First note that the determinant preserves multiplication. That is, if #A. B# are any matrices then:

#abs(A B) = abs(A) * abs(B)#

In particular:

#abs(A^2) = abs(A)^2#

In particular note that if #A# has real coefficients then #abs(A) in RR# and #abs(A)^2 >= 0#.

We can deduce that it is not possible for the determinant of a squared matrix over #RR# to be negative.

We can deduce that a matrix such as #((0, 1), (1, 0))# (which has negative determinant) is not the square of any matrix over #RR#.

Can it be the difference of two squared matrices?

Yes - for example:

#((1/sqrt(2), 1/sqrt(2)),(1/sqrt(2), 1/sqrt(2)))^2 - ((1,0),(0,1))^2 = ((1,1),(1,1))-((1,0),(0,1)) = ((0,1),(1,0))#

Is this possible for all square real matrices?

I don't know, but note that not every matrix over integers is the difference of two squared matrices over integers. For example, the #1xx1# matrix #(2)# is not the difference of two squared integer #1xx1# matrices.

Notes

I did wonder if you actually intended to ask whether the difference of squares identity holds for matrices - i.e.:

#A^2-B^2 = (A-B)(A+B)#

This does not hold in general due to matrix multiplication being non-commutative in general.

We have:

#(A-B)(A+B) = A^2+AB-BA-B^2#

So if #AB != BA# then #(A-B)(A+B) != A^2-B^2#