What are the quantum relations of four-particle GHZ states? How to represent a four particle GHZ states in quantum cryptography and what is the relation between them?
$\newcommand\ket[1]{\left|#1\right&gt;}$
Since you ask about 4-particle GHZ state, I’ll assume you’re familiar with the more usual 3-particle GHZ state. 

The 4-particle GHZ state is $(\ket{0000}+\ket{1111})/\sqrt2$. Like the 3-particle GHZ state, loosing a single particle transforms this state to a mere classical mixture of $000$ and $111$. 
The relation between the particles can be expressed by the stabilizer group generated by :
$$\begin{gather}
XXXX\\ZZII\\IZZI \\ IIZZ.
\end{gather}$$
This group contains


the  6 permutations of $ZZII$, meaning, that if you measure any two particle in the $Z$ direction, you will obtain the same result
$XXXX$ : meaning that if you measure all particle in the $X$ direction, you will have an even number of them in each direction (i.e. all in the same direction, or two in one direction, 2 in the other, but never 3 in one direction and 1 in the other)
$YYYY$ : meaning that if you measure all particle in the $Y$ direction, you will have an even number of them in each direction.
$ZZZZ$ : meaning that if you measure all particle in the $Z$ direction, you will have an even number of them in each direction. (This is consistent with 1.)
the 6 permutations of $-XXYY$ : if you measure two particle in the $X$ direction, and two in the $Y$ direction, you will have an odd number of $-1$ results.


I thought combining 2,3, and 5 leads to a contradiction similar to the GHZ paradox, but I didn’t manage to find the correct combination, except that measuring any of the qubit the in $X$ or $Y$ direction project the 3 others in a 3-particle GHZ state.