# Is this (conjectural) geometric property intrinsically related to the distribution of Primes?

Given the series of prime numbers greater than 9, we can 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.

The blue points represent the primes, whereas the red points represent the integers that are located in the rows $$1,3,7,9$$ but that are not primes.

For instance, in correspondence of the second ten ($$x$$-axis) we find two red points in the rows $$1$$ and $$7$$ ($$y$$-axis), because $$20+1=21$$ and $$20+7=27$$ are not primes.

I have conjectured that given any two red points, it is always possible to find an ellipse with foci in these two points and passing through at least one blue point (a prime number) and at least another red point (a composite number).

Here I show some examples:

Similarly, it can be conjectured that given any two blue points, it is always possible to find an ellipse with foci in these points and passing through at least one prime and at least one composite. Here follows some examples:

My question is:

If true, is this property related to the distribution of the prime numbers? Or it is only due to the particular reference system (lattice) I used to represent them?

Thanks for your suggestions and comments, and sorry if the whole problem may be naive.

• Could you clarify what "passing by" means? – DanielWainfleet Sep 23 '18 at 10:20
• @DanielWainfleet Sure, I mean passing through, I edit. Thanks! – user559615 Sep 23 '18 at 10:23
• It might be fruitful to try this using a different base, or scaling the coordinates by some integral factor. Then we would have more confidence it not being due to the choice of reference. – Couchy311 Sep 24 '18 at 6:49
• @Couchy311 Sure, both are good hints. I will do it! Thanks! – user559615 Sep 24 '18 at 7:00

I think this property is true but has nothing to do with ellipses. Given any two points $$A=(x_A,y_A)$$, $$B=(x_B,y_B)$$ (with $$y_A\ne y_B$$) in your grid, and their midpoint $$M$$, there are at least $$x_A+x_B-3$$ couples of points, in your grid, symmetric with respect to $$M$$: just take any point $$(x,y_A)$$ on the grid with the same ordinate as $$A$$ (with $$x\ne x_A$$ and $$0), and its symmetric $$(x_A+x_B-x,y_B)$$.

If $$y_A= y_B$$ there are, instead, at least $$3(x_A+x_B)-3$$ couples of points, symmetric with respect to line $$x=(x_A+x_B)/2$$.

Any such couple of points belongs to a suitable ellipse of foci $$A$$, $$B$$, but I wouldn't say this is a peculiar property of ellipses. And it is of course unlikely that all these couples be formed by two points of the same colour, at least for large enough values of $$x_A$$ and $$x_B$$.

In all the exposed examples the four points (two foci and the other two points) form a parallelogram with one of the two diagonals on the two foci. Also on a parallelogram you can always form two ellipses with spotlights on each diagonal and pass through the other two points.

So, it is enough to prove that there are parallelograms with the properties mentioned to prove that there are such ellipses. (The opposite is not true, there could be ellipses, not the examples, that meet these properties and whose four points do not form a parallelogram).

For a given diagonal, you can look for a point that is a prime number (blue), and see if there is "counterpoint" (does not leave the area) and that it is red (compound number), and with that you have another example.

Calling the four points: A, B, C, D. And being A and C the points of the diagonal of the parallelogram which are in turn the foci of the ellipse. Given point C, point D would be:

$$D = A + C - B$$

For example:

$$D = (3,3) + (8,7) - (7,1) = (4,9)$$

Which is the second ellipse of the first figure.

In addition, it can be seen as:

$$33 + 87 - 71 = 49$$

• Thanks for your answer. It looks very interesting! – user559615 Sep 25 '18 at 11:52