First published June 26, 2014 with major revisions and additions through August 12.
Special thanks to Shafiq U. Ahmed for his encouragement and advice.
A logical proof of the TwinPrime
Conjecture
"There are infinitely many
primes p such that p + 2 is also prime"
You don't need a team
of mathematicians to prove the twinprime
conjecture.
I provide the following proofs, fully cognizant that I use that term for myself only:
And, last but not least,
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All these attempted proofs share a statement of arithmetic fact, as follows:
If X^{2}  1 is semiprime then X  1 and X + 1 are its prime factors.
To unpack the meaning of this:
If X^{2}  1 is semiprime
(12 * 12)  1 = 143 then X  1 and X + 1 are its prime factors
11, 13
That is, P and P + 2.
Dear Reader, if you understand the arithmetic, please continue....
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A logical proof begins with two
sentences:
Statement 1
A natural number N of form X^21 where X>3
can be expressed as the product of two factors:
X1 and X+1.
Statement 2
Of the infinite set of Ns of a form X^21, there
is an infinite subset for which X1 and X+1 are
prime factors.
The first argument is an
arithmetic property that does not require
elaboration. It states that there is an infinite
set of square numbers minus one that are the
product of two factors, composite or prime.
The second argument states that there is an
infinite subset of square numbers minus one that
are the product of two prime factors (semiprime).
The second argument is correct using the following
logic: Of the set of X^21 there is a subset for
which both factors are prime. (To be clear, X1
and X+1 can also be composite  one or the other
or both. Although a deterministic proof of
Statement 2 cannot be provided yet, it is proved
by the Infinitude of Primes and the Fundamental
Theorem of Arithmetic. Using axiomatic set theory,
both the set and the subset must be infinite.
I know, I know: Statement 1 appears to be banal, and Statement 2 seems to be wishfulthinking.
What if I added a third statement just to be perfectly clear about how simple the distribution of twin primes is?
Statement 3
All Ps and P+2s are of the form  and only of the form  described in Statement 1.
However, I have learned that academic mathematicians need more proof than commonsense logic can
provide. This is because they are concerned about infinity. I will therefore attempt to prove in my naive fashion Statement 2 by induction and Statement 3 by algorithm.
The algorithms are interesting because they reveal the obvious truth that finding twin primes, by virtue of their multiplicative products (semiprimes) having a single position in relation to perfect squares and the quadratic interval, is subpolynomial (nearly linear actually).
This is followed by what I like to think is my best attempt so far: a proof in the style of Euclid's proof of the infinitude of primes. I do this in the forlorn hope that it will spur a truly mathematical mind to find the simple solution to this problem, rather than the most complicated and obscure!
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First published July 19,
2014 with subsequent revisions
To prove Statement 2, begin
with the following elaboration of Statement 1:
If a natural number of the
form X^21 is greater than (X1)^2 / 2 by a
prime and less than (X+1)^2 / 2 by a prime
then X1 and X+1 are twin primes.
What does
this really mean?
I will now provide a proof by
induction that of the infinite set of natural numbers
X1 and X+1 (which can be compositecomposite, compositeprime, primecomposite, or primeprime) there is an infinite subset of X1 and X+1 that are primeprime:
i.
If a natural number N of the form X^21 is an odd number, there
exists an evennumber difference with the
previous square number ("perfect square").
ii.
There is an infinitude of such Ns.
iii.
All prime numbers can be expressed as half an
even number.
iv.
All prime numbers can be expressed as half the
difference between an odd N of the form X^21 and the previous perfect
square.
Lemma
There are an infinitude of primes such that P =
X^2 +Y^2 (Fermat's theorem on the sums of two
squares), and there is an infinitude of primes
such that P = X^2 +Y^2 (Pythagorean primes)
where X = Y1.
v.
The Ns of Step IV are of the same form as the
subset of the Pythagorean primes where X=Y1.
vi.
It follows that there are an infinitude of
primes of which Step IV and Step V are true.
vii.
Thus, there must be an infinitude of Ns of
the form X^21 for which the factors X1 and X+1
are both prime.
Therefore, "There are
infinitely many primes p such that p + 2 is also
prime".
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First published July 29 (added "A la Fermat" August 12)
2014
To prove Statement 3, begin
with the following huge "cheats" for an algorithm:
1. You only need to look at a SINGLE pair of Ns for EACH quadratic interval.
2. This pair can only be in ONE place relative to the square of the even number between them.
From these premises we can demonstrate two distinctive methods: "Trivial division" or "A la Fermat".
This algorithm's efficiency exploits the easy factoring of X^2X*2. A fast primality test can be used when the first factor is not trivial.
This algorithm has nearlinear scalability  restricted only by the efficiency of a fast primality test, such as MillerRabin.
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The base case of the proof  the
elaborated Statement 1  says that the
X1 and X+1 factors can be one of the following
pairs (and in no particular order):
X1 is
composite, X+1 is prime
X1 is prime, X+1 is prime
X1 is prime, X+1 is composite
X1 is composite, X+1 is composite
The following four successive values
of X^21 illustrate the four possible
combinations of composite and prime, with red
being prime numbers:
(X^2)1

(X1)^2

(X+1)^2

(X^2)1
 (X1)^2

(X+1)^2  (X^2)1

X1

X+1

783

729

841

54

58

27

29

899

841

900

58

62

29

31

1023

961

1089

62

66

31

33

1155

1089

1369

66

70

33

35

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To step back, correctly
understood, Statement 1 is saying by
implication that the twinprime conjecture is a
verisimilitude of the Infinitude of Primes and
the Fundamental Theorem of Arithmetic.
That the
twinprime observation ever acquired the
sobriquet conjecture is, actually, something of
a mystery unto itself. Why it became established as
"one of the great open questions in number theory" most likely lies in
its very late "discovery", in the mid19th century.
One can imagine that Euclid 
if he was the first person to realize the Infinitude of Primes  as a geometer
saw that twin primes are an artifact of square numbers and would not have thought
it an observation worth making. If he had made the observation then, it would not have been the "mountain out of a molehill" that
modern mathematicians have made of it.
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In the following I will attempt to state the infinitude of twin primes as Euclid might have stated it (or at least in the same spirit).
Let's first agree on what the Proof of the Infinitude of Primes says using, according to the Prime Pages, wording "closer to that which Euclid wrote, but still using our modern concepts of numbers and proof". (I have changed P to X for clarity.)
Theorem
There are more primes than found in any finite list of primes.
Proof
Call the primes in our finite list p_{1}, p_{2}, ..., p_{r}. Let X be any common multiple of these primes plus one (for example, X =p_{1}p_{2}...p_{r}+1). Now X is either prime or it is not. If it is prime, then X is a prime that was not in our list. If X is not prime, then it is divisible by some prime, call it p. Notice p cannot be any of p_{1}, p_{2}, ..., p_{r}, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete.
Allow me now to make an equivalent statement for the Infinitude of Twin Primes:
Theory
There are more twin primes than found in any finite list of twin primes.
Suggested Proof
Call the primes in our finite list p_{1}, p_{2}, ..., p_{r} of which there are at least two of the form p and p+2. Let X be any number of the form N^21 that is the product of one or more of these primes. Now X is either the semiprime of p and p+2, or it is not. If X is not semiprime, then it is divisible by some factor, call it f, that is the square root of the next odd square number. When all the primes less than p and p+2 are exhausted there will be another N^21 that is p*p+2 where p+ 2 is the prime square root of the next odd perfect square. Either way, the original list of primes that are twinned was incomplete.
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First published July 31, 2014
I will now state a Euclideantype full Proof of the Twin Prime Conjecture as follows:
i.
Consider any finite list of square numbers ("perfect squares").
ii.
By the method of common differences we know that this growth is quadratic with a 2nd degree common difference.
iii.
Thus, a polynomial describing this growth contains as its first term a square X^{2}.
iv.
Therefore, the factors of the perfect squares ^{[a]} must also be growing by 2.
v.
Thus, we see that the highest factor of the composite immediately preceding an even perfect square is the square root of the next odd perfect square.
vi.
Further, we know that these common factors, prime or composite, are an infinite set whose members are every odd number.
vii.
Because we know that every odd number is a member of this set and because the common difference of quadratic growth is 2, for every composite of the form X^{2}1 the two largest factors (prime or composite) must be X1 and X+1. ^{[b]}
viii.
Therefore, when the next square (X + 1)^{2} is semiprime, its prime root is one prime factor of the previous square X^{2}1. ^{[b]}
ix.
Thus, when all the composites less than X^{2}1 are factored, X^{2}1 is the semiprime for which said prime root is one factor.
x.
Therefore, this X^{2}1 is the product of two prime factors, one less than X, which is X1, and one more than X, which is X+1.
xi.
Further, Steps VIII and IX must recur with the infinite occurrence of prime square roots ^{[c]} to produce an infinite set of semiprimes X^21.
Therefore, "There are infinitely many primes p such that p + 2 is also prime".
[a] The members of the set of odd natural numbers from 3 to the X of the given X^{2}.
[b] As proved by the Fundamental Theorem of Arithmetic
[c] As proved by the Infinitude of Primes
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To clarify Steps I, II, and V, take this example:
i.
Consider any finite list of square numbers ("perfect squares").
4, 9, 16, 25, 36, 49
ii.
By the method of common differences we know that this growth is quadratic.
4, 9, 16, 25, 36, 49
5, 7, 9, 11, 13
2, 2, 2, 2
(To quote and credit Purplemath: Finding the Next Number in a Sequence: The Method of Common Differences: "Since these values, the "second differences", are all the same value, then I can stop. It isn't important what the second difference is (in this case, "2"); what is important is that the second differences are the same, because this tells me that the polynomial for this sequence of values is a quadratic.")
v.
Thus, we see that the highest factor of the composite immediately preceding an even perfect square is the square root of the next odd perfect square.
99 = 3 * 3 * 11
100
121 = 11 * 11
143 = 11 * 13
144
169 = 13 * 13
195 = 3 * 5 * 13 = (13 * 15)
196
225 = 3 * 3 * 5 * 5 = (15 * 15)
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Still Don't Get It?
Let's try a picture...
The point of this is to illustrate the knot of numbers around the perfect squares that show the factors and their products (primesemiprime, red; nonprimecomposite, blue) that happen to be twinned (prime or composite). This is the basis for my algorithms.
Whereas a mathematician might think about the twin primes first, giving them primacy (after all, they are primes), I look at it from the perspective of the products first, using basic arithmetic and the unique factorization of every composite number. This is not an unreasonable point of view. Many fundamental number theorems, such as the granddaddy, Fermat's Little Theorem, exploit the properties of squares. Squares have properties that "square roots" do not have.
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Now what I've told you does
not seem to be all that earthshattering, does
it?
However, it turns out that
this is the whole story as to where you'll find
twin primes. The frequency is subject to the
rapid expansion of square numbers and the
gradual thinning of the prime numbers. The
distribution is subject to the coincident points
where the X1 and X+1 factors are also prime
numbers.
 We know that the perfect
squares are growing exponentially.
and
 We know that the frequency of the primes is
diminishing asymptotically.
We infer that these two
artifacts of the number line can cause the
appearance of an apparently chaotic coincidence
of the prime factors for square numbers minus 1.
This distribution suggests to those who assume
complexity without testing their premises that
proving the twinprime conjecture is difficult 
when it is, essentially, selfevident.
I suggest that solving the Rubik's cube is a
much harder proposition than this problem could
be in a multitude of arithmetic universes. Simple
things can appear complicated, and complicated
things can appear simple. This is most
certainly in the former category.
Your comments, suggestions,
objections, and counterarguments are most
welcome on Quora.
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There is one very nice
practical result of this proof that is
selfevidently correct and that also empirically
proves it. This is simply that you can
infinitely traverse the number line in
polynomial time (using only the operations of addition, subtraction, multiplication, and nonnegative integer exponents) to find an infinite number of
twinned primes (as illustrated in the Excel VBA code example).
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There is a ton of empirical
evidence on this site, since 2007, that
illustrates the general distribution of twin
primes  and twinned primes  is predicated on
and predicted by the geometric expansion of
perfect squares: the "quadratic interval".
In truth, the number line is
neither a line nor is it straight. Twinprime
and twinnedprime distribution is unbounded.
The architecture of numbers does not enter a
parallel universe, understood only by the
selfselected few, when they reach a certain
outlandish magnitude. We do not require our
mathematicians to count "how many angels can
dance on the point of a needle".
Let's begin with a thought
experiment... “Go
forth and multiply” .
  
     
     
     
     
   
Monty Hall Problem for Dummies
No other mathematical puzzle
has produced more heated arguments and more
misunderstanding than the Monty Hall Problem.
It's tempting to call it deceptively simple, but
the truth is it's deceptively complex. In fact,
I can simplify its essence to one question: If
you were handed two decks of cards, one with
2/3 aces and the other with 2/3 jokers, and
were asked to pick an ace, from which pack
would you draw the card? If
that is not a sufficient clue, you need to
read this.
Factor
Families
A factor family is a set of
composites that share common factors. The family
begins with a semiprime, and all members of the
set share the two prime factors of the
semiprime. The family can be extended
indefinitely. For example, for the first 100,000
odd composites the factor family of 4619
(149*31) looks like this:
There are a few things to note
about this representation: The prime factors (in
red) are distinct, so, for example, 3 can be 3*3
or 3*3*3, and so on. The complete factors
(nonprime and prime in blue) of each composite
all have three common factors: the semiprime and
its prime factors. By definition, a semiprime
cannot have nonprime factors.
Download Excel
worksheet: Factor
Families (1.5
MB)
Modal Distribution of
Distinct Prime Factors:
It’s Not Necessarily What You Might Think
Yes, you can find the answers
to life's most persistent questions here! What
is the cumulative distribution of distinct prime
factors? The answer is both what you were
expecting and more than you were expecting. As
expected, the cumulative frequency of unique
prime factors diminishes with size, so that 2 is
the most frequent factor, 3 is the secondmost
frequent factor, 5 the third, and so on for all
prime factors. However, it turns out that the
cumulative frequency of the largest
prime factors  the biggest factor of each
composite  has a modal distribution that is increasing
at a slow rate approximating the cube root of N.
Up to 10,000, this mode is 19. Up to 1,000,000,
this mode is 73. It's fairly safe to assume that
if we graph any given set of numbers, there will
be a distinctive peak near P^3 = N.
Distribution of the
largest prime factor for each composite less
than 10,000. (The difference
between the even and odd curves is accounted
for entirely by the exclusion of prime
numbers.)
Download Excel
worksheet: Distinct Factor Analysis
(macroenabled) (30KB)
Paired,
Squared, Cubed, and Primed...?
If we pick a square or a cube
and count up to the nearest prime number, what
are the odds that a prime number will also exist
if we count up by the same amount from its root?
The answer, it turns out, is surprisingly high 
and even higher for cubes than for squares. If
the question sounds confusing, it's easy to
illustrate:
Squared example:
Square 
Prime 
Offset 
Root 
Prime 
Offset 
3025 
3037 
+12 
55 
67 
+12 
Cubed example:
Cube 
Prime 
Offset 
Root 
Prime 
Offset 
79507 
79531 
+24 
43 
67 
+24 
The results are:
 For squared numbers up to 1
million, this relationship holds true 41.0%
of the time (410 out of 1,000).
 For cubed numbers up to 1
billion, this relationship holds true 59.8%
of the time (598 out of 1,000).
Download CSV files: Squared1000
(30KB) and Cubed1000
(35KB)
 For squared numbers up to
100 million, this relationship holds true 31.4%
of the time (3,143 out of 10,000).
 For cubed numbers up to 1
trillion, this relationship holds true 44.7%
of the time (4,472 out of 10,000).
Download CSV files: Squared10000
(351KB) and Cubed10000
(425KB)
What has 6
got to do with it?
Here is a little observation:
The difference between the first
and fourth prime number of a
proximateprime polynomial is
ALWAYS A MULTIPLE OF 6.
For example:
Prox. Prime Poly.

1st Term

4th Term

4t1t

7th Term

7t1t

10th Term

10t1t

13th Term

13t1t

n^2 + n
+ 10157

10159

10177

18

10213

54

10267

108

10339

180

n^2  n
+ 10331

10331

10343

12

10373

42

10421

90

10487

156

2n^2 
2n + 10627

10627

10651

24

10711

84

10807

180

10939

312

n^2
 n + 11777

11777

11789

12

11819

42

11867

90

11933

156

n^2  n
+ 12107

12107

12119

12

12149

42

12197

90

12263

156

2n^2 
2n + 12277

12277

12301

24

12361

84

12457

180

12589

312

2n^2 
2n + 12409

12409

12433

24

12493

84

12589

180

12721

312

3n^2 
3n + 12653

12653

12689

36

12779

126

12923

270

13121

468

n^2 + 5n
+ 12785

12791

12821

30

12869

78

12935

144

13019

228

n^2 + n
+ 12887

12889

12907

18

12943

54

12997

108

13069

180

Surprisingly, this divisibility
by 6 does not stop with the fourth term. It
recurs with the polynomial's 7th term, 10th
term, 13th term, and so on ad infinitum.
Lots more data available: PPPs
< 50000, t1  t15 ("T" denotes each t
value divisible by 6) Download (115KB)
Do
you know why?
Factoring in Polynomial
Time: A Pronic Solution...
Sometimes the best things
in life are free  well, almost free... and
very simple... and blindingly obvious. A single
GCD
calculation using the closest pronic
number to N will produce a factor for
onethird of composites not divisible by 2 or
5 up to any size. For example, the
nearest pronic to 898097881 is 898110992, and
these numbers share a GCD of 1873  a prime
factor of both numbers.
An analysis of N <
10,000,000 shows that 35.8% of the nonobvious
composites are factorable with a single GCD
calculation. What is the common characteristic
of this huge class of numbers? They appear to
conform to rational
angles in the Sacks number spiral. Such
numbers can be generated with RadiusTest
using the Lines option and produce polynomials
with a third coefficient of 0.
Expanding the GCD calculation to
pronic numbers within 6 quadratic intervals of N
provides a nearly instantaneous factorization
test for more than twothirds of composites
ending in 1, 3, 7, or 9 regardless of magnitude.
Here is an analysis of composites less than 10^{7}
with GCDs calculated for pronics from 6
quadratic intervals less than N through 6
quadratic intervals greater than N. It shows a
74.4% success rate.
Analysis
for N<10^{7} Download
(11MB)
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