Is there a $3\times 3$ magic square adding up to $7$?

Suppose we have a $3 \times 3$ magic square with integer entries, and each row/column/main diagonal adds to $n$:

$$\begin{pmatrix} a &b & c \\ d& e & f \\ g &h &i \end{pmatrix}$$

Then

\begin{align} 3n& =(a+b+c)+(d+e+f)+(g+h+i) \\ &=e+(a+i)+(b+h)+(c+g)+(d+f) \\ &=e+(n-e)+(n-e)+(n-e)+(n-e) \\ &=4n-3e \end{align}

Thus $n=3e$ so $n$ must be a multiple of $3$. Unfortunately, $7$ is not a multiple of $3$.