Let a magic square (row, columns and diagonals total the same amount) by a 3 by 3 matrix with entries $abc\ def\ ghi$. Let the polynomials inside the ideal all equal zero. Let ideal be generated by the polynomials $$I=\langle a+b+c-d-e-f,a+b+c-g-h-i, d+e+f-g-h-i, a+d+g-b-e-h, a+d+g-c-f-i, a+e+i-c-e-g, b+e+h-c-f-i \rangle \subset \mathbb{Q}[a,\ldots ,i].$$ Show that if $F \in I$, then $F$ is zero on any magic square.
Then show that $$(100a+10b+c)^2+(100d+10e+f)^2+(100g+10h+i)^2-(100c+10b+a)^2-(100f+10e+d)^2-(100i+10h+g)^2 \in I.$$
For the first part I know F would be a linear combination of the generators, but how do I show it zero?
For the second part I need to show it a linear combination of the generators. How would Groebner bases help me here?
Here is an image of a Maple session.
I used the symbol $U$ for the ideal generated by the linear polynomials corresponding to the equations based on the specification that your $3{\times}3$ matrix is a magic square. In your OP, you called it $I$, but in Maple, $I$ is a reserved symbol.
I used the symbol $p$ for the polynomial you called $F$ in your OP.
The goal is to show that $p \in U$.
Let $V = (U,p)$ be the ideal generated by $U$ together with $p$.
I computed $U$_bas, the Groebner basis for $U$, and $V$_bas, the Groebner basis for $V$. Since, as it turns out, the resulting Groebner bases are equal, it follows that $V=U$, hence $p \in U$, as was to be shown.