Let $$A$$ be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of $$A^2$$ is 1, then the possible number of such matrices is:

Let $$A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$$ be a symmetric matrix of order 2 with integer entries. Then $$A^2 = \begin{pmatrix} a^2 + b^2 & ab + bc \\ ab + bc & b^2 + c^2 \end{pmatrix}$$.

The sum of the diagonal elements (trace) of $$A^2$$ is $$a^2 + b^2 + b^2 + c^2 = a^2 + 2b^2 + c^2 = 1$$.

Since $$a, b, c$$ are integers and $$a^2 + 2b^2 + c^2 = 1$$ with all terms non-negative, we need $$2b^2 \leq 1$$. Since $$b$$ is an integer, this forces $$b = 0$$.

With $$b = 0$$, we need $$a^2 + c^2 = 1$$. The integer solutions are $$(a, c) \in \{(1, 0), (-1, 0), (0, 1), (0, -1)\}$$.

This gives exactly 4 symmetric matrices with integer entries whose squared trace equals 1.