If $A$ is an $n\times n$ integer matrix, then trivially $S=A+A^t$ and $P = AA^t$
where $t$ is ``transpose", are both symmetric.


Assume that $A$ is also a "$\lbrace -1,1 \rbrace$" matrix, i.e., the square of each entry in $A$ is equal to $1$.


Is there some rational-coefficient symmetric polynomial $P(x,y)$ (depending possibly on $A$ ?)
such that
$$
P(A,A^t)
$$
is also a $\lbrace -1,1 \rbrace$ matrix?



Symmetric polynomials preserving $-1,1$ matrices
As already mentioned in the comments above, it's not a big deal to find such a polynomial for a particular $\pm1$ matrix. If the question is about a "universal" polynomial (that is, depending only on $n$), then I would expect the answer "no". For $n=2$, take the matrix
$$
A=\left(\begin{matrix} 1 & -1 \cr 1 & 1 \end{matrix}\right).
$$
Then $A+A^t=2I\_2$ and $AA^t=2I\_2$ where $I\_2$ denotes the $2\times 2$ identity matrix. Because any symmetric polynomial with numerical (not matrix!) coefficients is a polynomial in $A+A^t$ and $AA^t$, its value will be always $cI\_2$ for a certain numerical constant $c$. Therefore, the off-diagonal entries will be always zero.


