How to Compute C^2 in AES MixColumns Matrix? In mix Columns we have:

$$C(x) = \{03\}X^3 + \{01\}X^2 + \{01\}X^1 + \{02\}$$

In Viktor Fischer's Paper on MixColumn and InvMixColumn Resource Sharing, in page 2, Equation 8, it's been said that $C^2(X)$ can be computed as follows:
$$C^2(x) = \{04\}X^2 + \{05\}$$ 
Can anybody explain why is that?
hi Ruby i know how it comes. But before that i assume you have a good knowledge of finite field.

Actually, it's the problem of how to calculate $$C^2(x) = (\{03\}X^3 + \{01\}X^2 + \{01\}X^1 + \{02\})^2$$. Becuase the character of $K$(as in the paper) is 2, so $$C^2(X)=(\{03\}X^3)^2+(\{01\}X^2)^2+(\{01\}X^1)^2+(\{02\})^2$$.  But you have to know the meaning of $\{03\}$  $\{02\}$ and $\{01\}$. They are elements of $F_2^8$. For example $\{03\}$ stands for $x+1$, so $(x+1)^2=x^2+1=\{05\}$. Like this we get:$$C^2(X)=\{05\}X^6+\{01\}X^4+\{01\}X^2+\{04\}$$
Next as $R=K[X]/(X^4+1)$, you have to put them in $R$ by  $mod(X^4+1)$$(X^j mod (X^4+1)=X^{j mod  4})$. so we get:$$C^2(X)=\{05\}X^2+\{01\}+\{01\}X^2+\{04\}$$. 
Next as $\{05\}=x^2+1,\{01\}=1,\{04\}=x^2$, so when we 'add'(this add in the finite field means XOR)then we get:$$C^2(x) = \{04\}X^2 + \{05\}$$