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.

I think I have a closed formula for the missing points. These seem to be (conjeture, but should not be difficult to prove) the composite integers $n\in\mathbb{N}$ such that $n=\pm 1 \mod 6$. This sequence gives: 1,25,35,49,55,65,77,85,91,95,115,119,121...