Let $M=\begin{pmatrix}a&b\\ b&c\end{pmatrix}$ be a $2\times 2$ symmetric matrix with integer entries.

For a $2\times 2$ matrix, invertibility is determined by its determinant: $\det(M)=ac-b^2$ So, $M$ is invertible iff $ac-b^2\ne 0.$

We now examine each option.

1. Option A

The first column of $M$ is $\begin{pmatrix}a\\ b\end{pmatrix}$ and the transpose of the second row is also $\begin{pmatrix}b\\ c\end{pmatrix}^T \text{ transposed }=\begin{pmatrix}b\\ c\end{pmatrix}.$ But the statement says first column equals transpose of second row, so $\begin{pmatrix}a\\ b\end{pmatrix}=\begin{pmatrix}b\\ c\end{pmatrix}$ which gives $a=b,\quad b=c.$ Hence $a=b=c,$ so $M=\begin{pmatrix}a&a\\ a&a\end{pmatrix}.$ Then $\det(M)=a^2-a^2=0.$ So such a matrix is not invertible.

Therefore, A is false.

2. Option B

The second row of $M$ is $\begin{pmatrix}b&c\end{pmatrix}$ and the transpose of the first column is $\begin{pmatrix}a\\ b\end{pmatrix}^T=\begin{pmatrix}a&b\end{pmatrix}.$ Equating them: $\begin{pmatrix}b&c\end{pmatrix}=\begin{pmatrix}a&b\end{pmatrix}$ so $b=a,\quad c=b.$ Thus again $a=b=c,$ and $M=\begin{pmatrix}a&a\\ a&a\end{pmatrix}.$ Hence $\det(M)=0.$ So $M$ is not invertible.

Therefore, B is false.

3. Option C

If $M$ is diagonal with non-zero entries on the main diagonal, then $M=\begin{pmatrix}a&0\\0&c\end{pmatrix}, \quad a,c\in \mathbb Z,\ a\ne 0,\ c\ne 0.$ Then $\det(M)=ac.$ Since both $a$ and $c$ are non-zero, $ac\ne 0.$ Therefore $M$ is invertible.

So, C is true.

4. Option D

The product of entries in the main diagonal is $ac$. Given that $ac$ is not the square of an integer.

Now if $M$ were not invertible, then $ac-b^2=0 \implies ac=b^2.$ But $b$ is an integer, so $b^2$ is the square of an integer. That would mean $ac$ is a square of an integer.

This contradicts the given condition that $ac$ is not the square of an integer.

Hence $ac-b^2\ne 0$, so $M$ must be invertible.

Therefore, D is true.

Final answer

The correct options are: $\boxed{\text{C, D}}$

This matches the stored correct answer.