The Strassen Algorithm for computing $AB$ where $A$ and $B$ are two even matrices involves splitting the matrices into submatrices and then reducing the number of multiplications by $1$ from $8$ to $7$ through some magic. I'd like to reduce the number of multiplications from $7$ to $5$ in the case of squaring a matrix: $AA$. I understand that after the first step of the algorithm, when I do the recursive call, this doesn't actually work because I will no longer be squaring matrices, however, I'd like to prove that it is possible to do so.
Starting from the Strassen Algorithm, you have:
$M_1 = (A+C)(A+B)$
$M_2 = (B+D)(C+D)$
$M_3 = (A-D)(A+D)$
$M_4 = D(C-A)$
$M_5 = (A+B)D$
$M_6 = (C+D)A$
$M_7 = A(B-D)$
Where $A,B,C,D$ are the sub-matrices. In an attempt to look to combine some of these equations, it seems like $1$ and $5$ could be combined, and $2$ and $6$ as well. However, the order of the multiplications doesn't work, and even if it did, I wouldn't be able to get some of the other products.
Is there anything obvious that I'm missing? (Perhaps "obvious" isn't the right word)