Groebner Basis question (ideals) 
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?
 A: "The equations inside the ideal all equal zero" is confusing.  You have a certain ideal $I$ of polynomials in indeterminates $a,b,\ldots,i$ (which for convenience I'll write as $X_1, \ldots, X_9$).   There are no "equations inside the ideal", just polynomials, and only one of them is $0$.  If this is over a field $k$ (perhaps $\mathbb C$ in your case), then the polynomials can be mapped to functions on $k^9$, and
there is the affine variety
$V(I)$ which is the set of $9$-tuples $(x_1, \ldots, x_9)$ such that
$F(x_1,\ldots,x_9) = 0$ for all $F \in I$.  I think the first question is to show that $(x_1, \ldots, x_9) \in V(I)$ if
$F(x_1,\ldots,x_9) = 0$ for each of the generators of $I$.
The second question is wrong, unless you're working over a field of characteristic $2$, $3$ or $11$.  For example, $$[a=-1, b=2, c=0,d=2,e=-1,f=0,g=0,h=0,i=1]$$
makes all your generators $0$, but  $$\left( 100\,a+10\,b+c \right) ^{2}+ \left( 100\,d+10\,e+f \right) ^{2
}+ \left( 100\,g+10\,h+i \right) ^{2}- \left( 100\,c+10\,b+a \right) ^
{2}- \left( 100\,f+10\,e+d \right) ^{2}- \left( 100\,i+10\,h+g
 \right) ^{2}
=  32076 $$
A: 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.
