# Thread: A method to show there are infintely many Twin Primes

1. ## A method to show there are infintely many Twin Primes

This is an attempt at a Twin Prime Conjecture Proof, and is open to Peer Review.

Abstract: Surfaces representing the primes and composites, the upper twin prime, the difference between squares, and the relations among them, are used to select input for a quadratic in such a way as to always generate a Twin Prime. Because each input generates a unique Twin, and because the input set is infinite, there are infinitely many twin primes.

Main: Begin with the first quadrant of the surface $z=xy$. Along the slices $x=1$ and $y=1$, the surface returns the counting numbers, and over the rest of the field, where $x\geq 2$ and $y\geq 2$, it returns the composites.

$z=xy$

Checking a specific value of z for non prime integer solutions shows whether that value is prime or composite. Accomplish this by first treating each height z as a constant, and then solve for y. This yields $y=\frac{z}{x}$.

$y=\frac{z}{x}$

Next, the line $y=-x+b$ is moved across $y=\frac{z}{x}$ for $4\leq b\leq \frac{z}{2}+2$.

$y=-x+b$

The lower limit on b ensures the (1,1) answer is ignored, and the upper limit comes from there being no factors greater than 1/2 of a number. Set the equations equal and solve for x. If there are integer solutions within b's range, then z is composite, if there are no integer solutions it is prime.

$\frac{z}{x}=-x+b$

$x^{2}-bx+z=0$

$x=\frac{b\pm \sqrt{b^{2}-4z}}{2}$

Allow $z+2=xy$ to represent the surface that will check 2 greater than a number, in this case, the 2nd prime of a pair. Solving for y and repeating the process gives:

$z+2=xy$

$y=\frac{z+2}{x}$

$x=\frac{b\pm \sqrt{b^{2}-4z-8}}{2}$

For both the Prime-Composite and Twin Prime surfaces, since b is an integer, x can only be an integer when the square root evaluates to a whole number. This means that $4z$ and $4z-8$ cannot be equal to a difference between squares, otherwise x runs a chance of being composite. Note, there are values b and z where x evaluates to a non integer multiple of 1/2 that are safe to use, but simply allowing the square root to not be whole is enough to ensure no integer solutions will exist.

$4z\neq x^{2}-\left ( x-y \right )^{2}$

$y^{2}-2xy+4z=0$

$y=x\pm \sqrt{x^{2}-4z}$

Again, a scanning line is used to check and ensure that y does not evaluate to a whole number. The new requirement due to the square root becomes:

$\sqrt{x^{2}-4z}\neq -x+b$

Solving for z yields:

$z=\frac{-b^{2}+2xb}{4}$

This is just another surface, however, the letters being used as variables at this point are not following the z(x,y) convention. For familiarity and personal preference, I rewrite the current surface in terms of x and y, where b becomes x, and x becomes y. This is simply a syntactical rewrite.

$z=\frac{-x^{2}+2yx}{4}$

Repeating the process for the Twin Prime surface gives:

$z=\frac{-x^{2}+2yx+8}{4}$

Initially, scanning a height z to check for primality, corresponds to a range of x, and has a restriction such that $1<x<z$, otherwise, products of 1 and a number contaminate the field. Checking larger z increases the range of x. Practically speaking, the minimum integer value of x needed to ensure the scan will cross the function is $\left \lfloor \sqrt{z} \right \rfloor = x$. At this point, if z takes any value from these surfaces for a corresponding range, which is now controlled by the variable y, then the previous quantity will be a difference of squares, and the original value will generate composites. Note, we can always check a larger z, such that our range's upper bound stops on any given y. Charting $z=\frac{-x^{2}+2yx}{4}$ with rows x and columns y shows the squares down the diagonal, due to its nature, and marks where the output of the chart repeats or becomes negative. Therefore, only values where $1\leq x\leq y$ need be considered.

The goal is to find values not appearing on either surface over a given range of y. It would be preferred to scan a range and collect all missing values, however, only one is needed. Any given column y, which by default corresponds to a maximum of a range, will always have values missing that are $> \left ( y-1 \right )^{2}$ that do not and cannot appear in any smaller columns. Furthermore, since the transformed Twin Surface is 2 above the Prime-Composite Surface, there will always be values missing at the bottom of its rightmost column that are not on the Prime-Composite surface. Noting that even values of y generate composites allows the search to be restricted to odd values of y only. Then, the missing number with no integer solutions on either surface is found by adding 3/4 to the lower surface or subtracting 5/4 from the higher. This gives:

$z=\frac{-x^{2}+2yx}{4}+\frac{3}{4}$

Since the value is coming from the surface where $y=x$, this simplifies the equation into the parabola for odd x.

$z=\frac{x^{2}+3}{4}$

This set will now contain only primes and some remaining odd composites for odd x. The next step is to filter these composites by writing them as the product of 2 odd numbers.

$\frac{x^{2}+3}{4} \neq \left ( 2n+1 \right )\left ( 2m+1 \right )= 4nm+2n+2m+1$

Which leads to the condition that:

$x\neq \sqrt{16nm+8n+8m+1}$

Removing those values as input leaves only primes as the output. The final step is to filter the regular primes from the twins. If an odd value is a twin, then 2 less than it will also be prime, however, if it is not a twin, then 2 less than it will be an odd composite. That is:

$\frac{x^{2}+3}{4}-2 \neq \left ( 2n+1 \right )\left ( 2m+1 \right )= 4nm+2n+2m+1$

Which gives the condition that:

$x\neq \sqrt{16nm+8n+8m+9}$

Therefore, when x is a value not on the surfaces $z\neq \frac{-x^{2}+2yx}{4}$ and $z\neq \frac{-x^{2}+2yx+8}{4}$, but is not a value from the surfaces $z\neq \sqrt{16nm+8n+8m+1}$ and $z\neq \sqrt{16nm+8n+8m+9}$, then $\frac{x^{2}+3}{4}=\textrm{A Twin Prime for odd x}$

In conclusion, showing that $\frac{x^{2}+3}{4}$ generates only primes, twin primes, and odd composites for odd x, and then removing the odd composite and non-twin prime generating inputs, proves the existence of infinitely many Twin Primes. It should follow, to replace the $z+2$ with $z+2n$ and show that the method can be used for any even prime gap.

Supporting pictures for this method can be found here: art-and-culture-blog

2. ## Re: A method to show there are infintely many Twin Primes

*****This is an edit in an attempt to make my question and the crux of this method clearer.

??? How does showing that there are infinitely many integer values that are not on $z=xy$
and $z+2=xy$, with x and y greater than 1, fail to prove the Twin Conjecture???

Numbers not on $z=xy$ are prime. Numbers not on $z+2=xy$ are a number 2 greater than a number that is prime. Numbers not on either surface, have a number 2 greater that is prime, and are themselves prime. These are the Twins. The task then becomes to show that there is an infinite set not on both surfaces. That’s it, it’s that simple.

Now I can’t do it directly from the raw data of the surfaces, but I can do it using a transform. By parsing the surface with a line, I set up a transform condition. The transform condition sets up a rule, and I use the rule to find an infinite amount of twins not on either surface. However, there is a mapping in the process. The transformed surfaces are such that the values they do take correspond 1:1 to all the composites and “number-2-away” composites, but the values they don’t take are not 1:1 to the twins.

There are extraneous fraction, odd composite, and non twin generating values in the gap. Since there are still endless numbers left once those are filtered, you have infinite twins.*****

3. ## Re: A method to show there are infintely many Twin Primes

**(Update)**

Given there are infinite primes.

Given $z\neq xy$ is the set of primes for $1<x$ and $1<y$.

Therefore, there are infinite holes, values with no integer solutions, on that surface.

Given there are an infinite number of primes 2 away from some number.

Given $z\neq xy-2$ checks the primality of a number $z+2$ for $1<x$ and $1<y$. This is the first surface slid down by 2.

Therefore, there are infinite holes on the second surface.

The Conjecture is now stated as to ask if there are infinite elements in the intersection of those 2 infinite sets.

Given if there is an integer x, such that $x=\frac{b\pm \sqrt{b^{2}-4z}}{2}$ for $2\sqrt{z}\leq b\leq \left \lfloor \left ( \frac{z}{2}+2 \right ) \right \rfloor$, then z is composite.

For the second surface, $2\sqrt{z+2}\leq b\leq \left \lfloor \left ( \frac{z+2}{2}+2 \right ) \right \rfloor$.

Therefore, there are an infinite number of solutions where x is not an integer.

Therefore, $b\pm \sqrt{b^{2}-4z}\neq even$.

Therefore, when b is even the square root must be odd, and when b is odd the square root must be even.

Therefore, the square root must not equal an integer.

Therefore, $4z$ and $4z-8$ must not equal a difference of squares.

Therefore, there are infinite spots where $4z$ and $4z-8$ do not equal a difference of squares respectively.

Given a difference of squares $x^{2}-\left ( x-y \right )^{2}$.

Therefore, $y\neq x\pm \sqrt{x^{2}-4z}$.

Therefore, $\sqrt{x^{2}-4z}\neq -x+b$ for $\sqrt{4z}\leq b\leq \infty$.

Therefore, $z\neq \frac{-b^{2}+2xb}{4}$, which I like to rewrite for convention as $z\neq \frac{-x^{2}+2yx}{4}$, has infinite solutions.

For surface 2, $z\neq \frac{-x^{2}+2yx+8}{4}$.

Therefore these surfaces represent the conditions to be checked.

Therefore, all values on these surfaces do not meet the conditions.

Therefore there are infinite values not on these surfaces that meet each's condition respectively.

Given, not all values not on these surfaces meet the conditions.

The Conjecture is now stated as to ask if infinite values, not on the surfaces, within the restrictions of the parameters, and that do meet both conditions, can be found.

Therefore for any odd x, $\frac {x^{2}+3}{4}$ will create a value that is not on either surface for the appropriate range of parameters.

Because there are infinite odd numbers, the parabola $\frac {x^{2}+3}{4}$ generates infinite such values.

Given the values are all odd composites, prime, or the second prime of a twin pair.

Given an odd composite $\left ( 2n+1 \right )\left ( 2m+1 \right )= 4nm+2n+2m+1$, then $\frac{x^{2}+3}{4} \neq \left ( 2n+1 \right )\left ( 2m+1 \right )= 4nm+2n+2m+1$.

Therefore, $x\neq \sqrt{16nm+8n+8m+1}$

Given that 2 less than a prime that is not a second twin is composite yields: $\frac{x^{2}+3}{4}-2 \neq \left ( 2n+1 \right )\left ( 2m+1 \right )= 4nm+2n+2m+1$.

Therefore, $x\neq \sqrt{16nm+8n+8m+9}$.

Therefore, as long as $x\sqrt{16nm+8n+8m+1}$ and $x\sqrt{16nm+8n+8m+9}$ do not sweep out the set of all odds, which they do not, then there are infinite values that meet both conditions.

Therefore, there are infinite Twin Primes. **(end edit)**

4. ## Re: A method to show there are infintely many Twin Primes

You keep talking about a surface. If z = xy and x and y are real, then z is a surface. But if x and y are positive integers, z is not a surface at all. So when you then talk about lines intersecting the surface, I am already lost.