The second condition is satisfied by sets such as f 1 ( x ) = x + 4 , f 2 ( x ) = x + 7 {\displaystyle f_{1}(x)=x+4,f_{2}(x)=x+7} , since ( x + 4 ) ( x + 7 ) {\displaystyle (x+4)(x+7)} is always divisible by 2. It is easy to see that this condition prevents the first condition from being true. Schinzel's hypothesis essentially claims that condition 2 is the only way condition 1 can fail to hold.