A conjecture involving prime numbers and circles

Question

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 through at least other 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 1

COMMENT.- It seems to me it is impossible to prove this geometrically. However, algebraically it could be maybe possible. The general equation of a circumference $\Gamma$ is $$x^2+y^2+Dx+Ey+F=0$$ and the compatibility condition for four points $(p_1,0),(p_2,0),(p_3,k),(p_4,k)\space$in $\Gamma$ is

$$\det\begin{vmatrix} p_1 & 0 & 1 & p_1^2 \\ p_2 & 0 & 1 & p_2^2 \\ p_3 & k & 1 & p_3^2+k^2 \\ p_4 & k & 1 & p_4^2 +k^2 \notag \end{vmatrix}=0$$

In this case $p_1,p_2,p_3$ are primes, the two first points are in the x-axis and the other in the line $y=k\ne0$.

Puting, for example, $p_1=37$ and $p_2=47$, the choice of $(p_3,k)$ determine for $p_3$ fixed (say $23$) and $k$ as parameter a family of circumferences $\Gamma_k$ defined by a quadratic equation (cubic?) $$Q(p_4,k)=0$$ in which the arbitrary variation of $k$ could give a prime $p_4$ maybe.

Answer 2

I'm not a mathematician, so take everything here with a grain of salt. I couldn't help but be interested though.

You can circumscribe a circle on any trapezoid. Any two primes equivalent mod $10$ can form one base of a trapezoid; any two primes not equivalent mod $10$ can form the side of a trapezoid, including primes $10x+y_1$ and $10x+y_2$, which would form a rectangle, special case of a trapezoid.

I'm going to treat the second case first, as I think it's actually easier. Let's take two random primes that have a difference of $10a-b$. If we can two primes with a difference of $10(a-1)+b$, with the same $y$ coordinates, then we can form a trapezoid. So this is basically equivalent to the conjecture that for every even $d$, there are primes $p>q$ for which $p-q=d$. This is an open conjecture which, if I'm not mistaken, is true if both the Goldbach conjecture and the twin prime conjecture are true.

For the first case, in order to find a trapezoid, we need a second pair of primes whose average is within $\pm 4$ of the average of the first pair of primes. So, can we find two primes with an arbitrary average? This is just the Goldbach conjecture: for any even $k$, there are primes $p, q $ for which $p+q =k$.

Answer 3

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

$$ \left\lfloor{ \frac{a}{10}} \right\rfloor - \left\lfloor{\frac{b}{10}}\right\rfloor = -\left(\left\lfloor{\frac{c}{10}}\right\rfloor - \left\lfloor{\frac{d}{10}}\right\rfloor\right) $$

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

$$ a = b,\; \; c=d \; \mod 10 $$

($a$ 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 $a = b \mod 10 \;$, the above argument still probably holds, but I have not found a proof.

Answer 4

Here is some loose intuition to convince you that it is equally hard as the twin prime conjecture. Especially, to convince you that there's no point in trying to prove or disprove it:

• At most as hard as the twin prime conjecture:
  Take two primes $p_1,p_2$. If the twin prime conjecture is true, it is reasonable to expect that, for any even $2k \geq 2$ and $n \bmod 10$ there are infinitely many prime pairs $(q_1,q_2)$ with $q_2-q_1 = 2k$ and $q_1 \equiv n \pmod{10}$.¹ Then for any given $p_1,p_2$ not congruent mod $10$ we can find two other primes to form a trapezium. This takes care of the case where $p_1,p_2$ are not congruent, at least.

• At least as hard as the twin prime conjecture:
  Four points with coordinates $(x_i,y_i)$ are concyclic iff $$\begin{vmatrix} 1 & x_1 & y_1 & x_1^2 + y_1^2 \\ 1 & x_2 & y_2 & x_2^2 + y_2^2 \\ 1 & x_3 & y_3 & x_3^2 + y_3^2 \\ 1 & x_4 & y_4 & x_4^2 + y_4^2 \end{vmatrix} = 0$$ This gives, for every pair of primes $(p_1,p_2)$ a degree $4$ equation in two primes $q_1,q_2$ (and their residues mod $10$). Current methods are nowhere near proving that it has a solution; indeed, we cannot even show that the degree 1 (!) equation $$q_2-q_1-2k = 0$$ has a solution for every $k$.

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¹ Although, there was an article that appeared a few years ago with some computations, suggesting that the distribution of the remainder of three consecutive primes mod a given integer, is not uniform. Anyway.