A conjecture involving prime numbers and parallelograms

The eight points on each parallelogram cover all residues mod 7.

This is an expansion of Michal's answer. We can find a multiple of 7 on each of the first seven parallelograms: 49, 77, 119, 91, 133, 161, 203.

Now every parallelogram can be obtained by taking one of these first seven parallelograms and adding 210 to each point. Since 210 is a multiple of 7, each point listed above will be translated onto missing points. This gives a way to compute a missing point on every parallelogram: if you find the remainder mod 210 of the points on the parallelogram, it must contain one of the seven values I listed above.