Given the series of prime numbers greater than 9, we organize them in four rows, according to their last digit (1,3,7 or 9). The column in which they are displayed is the ten to which they belong, as illustrated in the following scheme.

My conjecture is:

Given any two primes (i.e. given any two points in the above scheme), it is always possible to find a circle passing throughat leastother two points, representing other two primes.Here I present some examples, taking two random points. Sorry for the bad quality of the picture.

Since I am not an expert of prime numbers, this can be an obvious result (if true, of course). In this case, I apologize for the trivial question.

However, I tried to attack the problem by means of the equation of the circle, but I got lost. Thanks for your help!

NOTE: You might be interested in this and in this other post. Also, here I state a similar conjecture for ellipses.

**Answer**

Assuming Polignac’s conjecture we will always be able to find two primes (c,d) such that

⌊a10⌋−⌊b10⌋=−(⌊c10⌋−⌊d10⌋)

(the distance between a and b along the x-axis is equal to the negative of the distance between c and d) and

a=b,c=dmod10

(a and b, and c and d, end in the same digits).

This defines an isosceles trapezium, which is always a cyclic quadrilateral (a quadrilateral such that a circle can be drawn with its 4 vertices.

If a=bmod10, the above argument still probably holds, but I have not found a proof.

**Attribution***Source : Link , Question Author : Community , Answer Author : Direwolf202*