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.

$\displaystyle z=xy$

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

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

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

$\displaystyle y=-x+b$

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.

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

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

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

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:

$\displaystyle z+2=xy$

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

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

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.

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

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

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

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:

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

Solving for z yields:

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

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.

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

Repeating the process for the Twin Prime surface gives:

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

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:

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

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

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

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.

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

Which leads to the condition that:

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

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:

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

Which gives the condition that:

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

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