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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.

enter image description here

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.

enter image description here

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.

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    $\begingroup$ Fascinating...brilliant question! $\endgroup$ – Don Thousand Aug 17 '18 at 18:06
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    $\begingroup$ At first sight your conjecture is very beautiful. However it seems that, in order to make it plausible you must add some minimum "distance" between the two arbitrary primes when starting because a "little" distance could be such that any other prime is touch by the corresponding circle. (sorry for bad English) $\endgroup$ – Piquito Aug 17 '18 at 18:18
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    $\begingroup$ My comment was wrong in its very base because there are infinitely many circles passing by two points no matter what near are. Good luck. $\endgroup$ – Piquito Aug 17 '18 at 20:38
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    $\begingroup$ @LoganToll Sure: First I randomly choose 2 points, then I look for other 2 points from which a circle can pass. That's all. The point of the conjecture is that one can always find such circle, independently on the initial choice. $\endgroup$ – user559615 Aug 17 '18 at 23:34
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    $\begingroup$ then a question immediately pops out: what happens if the representation is done in a base other than 10. $\endgroup$ – G Cab Aug 18 '18 at 9:18
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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.

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    $\begingroup$ Thanks for your contribution. Just, I don't get the first point: I am not sure that the fact that the primes are infinite can ensure that your formula applies. Can you deepen this point? Thanks again! $\endgroup$ – user559615 Aug 19 '18 at 21:05
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    $\begingroup$ It follows trivially from Polignac's conjecture, which is probably true, but otherwise, I can't find any obvious way to prove it. Much like Polignac's conjecture (a generalisation of the twin prime conjecture) it is likely easy to state, and incredibly difficult to prove. $\endgroup$ – Direwolf202 Aug 19 '18 at 21:44
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    $\begingroup$ You should edit your answer to make it clear that it is just a conjecture. The way it is currently written, it looks like you are claiming a proof. $\endgroup$ – TonyK Aug 20 '18 at 13:57
  • $\begingroup$ TonyK - Clarified that it assumes Polignac's conjecture. I am claiming a proof, it is just dependent upon a different conjecture. Jyrki Lahtonen, good point, I didn't think through that properly, fixed. $\endgroup$ – Direwolf202 Aug 20 '18 at 14:54
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    $\begingroup$ If the two given primes are on the same horizontal line (i.e. are congruent mod 10), and if one of the primes is quite far to the left (the diagram does not continue to the left; if negative primes were included, things would be easier; the asker said primes greater than $7$), does your argument work? For example if one prime is $11$, at $(1,1)$, and the other prime is $131$, at $(13, 1)$, how can we proceed? I tried to find a "bad" example of this kind with a computer search but could not locate one. So maybe the conjecture is true for these cases also. $\endgroup$ – Jeppe Stig Nielsen Aug 21 '18 at 16:21
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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 $k$.


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.

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    $\begingroup$ Thanks for your answer, Barto. Yes, I agree with you. Other answers too go in the same direction. However, I think it is interesting to see the Goldbach and the Twins conjectures somehow gathered in only one, if I haven't misunderstood the comments to this post. I am new to the study of prime numbers and my ideas can be really naive. I apologize in this case, and thank you again for your answer! $\endgroup$ – user559615 Aug 20 '18 at 14:30
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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.

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  • $\begingroup$ I think your notation is different from OP's: OP designates primes by a tuple $(a,b)$ which represents the prime $10\times a+b$, and so $a$ and $b$ need not be primes. Also, on the $x$-axis there are no primes: it is divisible by $10$. $\endgroup$ – awllower Aug 18 '18 at 3:34
  • $\begingroup$ This is not properly the problem but a way of see how a circumference passing for three (arbitrary) points can have too another desirable point according with manipulating $p_4$ and $k$ in the quadratic $Q(p_4,k)=0$. (bad English, I know). $\endgroup$ – Piquito Aug 18 '18 at 3:49
  • $\begingroup$ @Piquito It seems to me that the conjecture is true for any conic (i.e. given any 4 random primes, we can always find at least other 2 primes belonging to the same conic). Can you check this? Thanks for your interesting comments! $\endgroup$ – user559615 Aug 18 '18 at 5:28
  • $\begingroup$ All depends of the corresponding $Q(p_4,k)$ above, I guess: fixing $p_4$ you do have freedom for $k$ so you plausibly can extend your conjecture (in case it is true). $\endgroup$ – Piquito Aug 18 '18 at 12:17
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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 $p+q =k$.

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  • $\begingroup$ Thanks for your answer and your observations. They look interesting. I am not a mathematician as well, so I have to study a bit more the conjectures you mentioned. Thanks again! $\endgroup$ – user559615 Aug 19 '18 at 4:04
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    $\begingroup$ Thankfully those conjectures are easily understood! In simple language, the Goldbach conjecture states that even even number is the sum of two (odd) prime numbers. The twin prime conjecture states that there are infinitely many "twin primes," that is, pairs separated by 2, like 3 and 5 or 41 and 43. $\endgroup$ – Eric Snyder Aug 19 '18 at 6:43
  • $\begingroup$ I see. So, you mean that to prove my conjecture would imply to prove both the "Goldbach" and "Twins" ones, don't you? $\endgroup$ – user559615 Aug 19 '18 at 6:55
  • $\begingroup$ I also realized there are a couple of types I've missed. There are some rectangles that are tilted with respect to Cartesian coordinates, that produce circles from primes on all four lines. The quartets (7, 29, 61, 83) and (13, 31, 79, 97) and (23, 59, 61, 97) are examples of those. There will also be sets of two on a vertical line and two on a horizontal line that will form trapezoids, for instance (23, 29, 41, 101) and (13, 17, 31, 71). All of these cases rely on finding similar triangles though, so I think they'll be uncommon past the small numbers. $\endgroup$ – Eric Snyder Aug 19 '18 at 7:03
  • $\begingroup$ What do you mean by "trapezoid"? Isosceles trapezia ($AB||DC$ and $AD=BC$) are cyclic but other trapezia are not. $\endgroup$ – Rosie F Feb 18 at 7:05

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