Suppose the opposite case. Take the sets $ A=(a,b,c,d)$ and $ B=(x,y,z,t)$. In each one of this ones, there is at most one even number, otherwise one of the modulus would be even. If all the elements in some set are odd, there would be a even modulus made by two odd elements from the other set. So each set has just one even element. But, in this case, is easy to see that some modulus will be even (you can suppose WLOG that some element of a set is even and divide in some cases), contradiction. It finishes the problem.