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Draft proof

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Draft proof

In the next few days I propose to develop a proof of the
following conjecture:

There are infinitely many primes of the form x^2 + 1 (where x belongs to N ).

Since I propose to use the tool entitled
"failure functions " I suggest that those interested may go to the site: Devaraj123.blogspot.com and read the piece on failure functions.


I have started this topic in this forum so that if
there is any error of logic or computation it can be
ponted out by members.
Each failure function covers a set of values of x. For
example,x = 1 + 2*k covers all the odd numbers in any
given interval. All these values of x, when substituted in f(x), f(x) is rendered a failure (composite ); in this case even numbers. Next: The failure function x = 2 + 5*k generates values of x such that f(x) are failures (composites), being multiples of 5.
Now let P be the largest prime of the form x^2 + 1 and let x =X be the relevant value of x . i.e. X^2 + 1 = P. Consider the interval X, X + P. There are two possible cases. Case I: all the values of x in this interval are covered by prior failure functions. Case II: There is atleast one value of x in the interval not covered by the prior failure functions. The first f.f. covers all odd numbers in the interval which need not be further considered as they will render f(x) even.
(To be continued )

What are the implications of case I? i) No prime greater than P is generated by f(x). ii) All the output of f(x) subsequent to x = X are composites being powers of some primes less than P. Here it is important to note that every polynomial has a large set (possibly infinite ) set of impossible prime factors.
f(x) has the following sequence of impossible prime factors: 3, 7, 11, 19, 23,.......
iii) Every f(x) subsequent to x =X is that the shape
of f(x) = 2^i5^j13^k..... where the exponents i,j,k... belong to W . iv) as x increases f(x) increases with only the exponents increasing but the base consists
of possible prime factors less than P.

Case II
What are the implications of case II?
i) If the failure functions generated prior to x = X do not cover all the values of x in the interval x, x+P each uncovered value of x, when substituted in f(x),
yields only a prime number which need not be tested for primality.
(to be continued )

Note that any value of x in the interval X, X + P which is not covered by the preceding failure functions, when substituted in f(x) yields a prime greater than P. Hence we will be designating by X the largest value of x in the above interval not covered by the preceding failure functions and the relevant value of f(x) by P. Also note that as a consequence the interval X, X + P becomes larger by this iteration.
I propose to show shortly that as the interval X, X + P grows larger the decrease in the fraction of the interval not covered by the preceding failure functions decreases only marginally leading to the conclusion that there are always a few uncovered values of x proving that the number of primes of form x^2 + 1 are infinetely many.

Note that any value of x in the interval X, X + P which is not covered by the preceding failure functions, when substituted in f(x) yields a prime greater than P. Hence we will be designating by X the largest value of x in the above interval not covered by the preceding failure functions and the relevant value of f(x) by P. Also note that as a consequence the interval X, X + P becomes larger by this iteration.
I propose to show shortly that as the interval X, X + P grows larger the decrease in the fraction of the interval not covered by the preceding failure functions decreases only marginally leading to the conclusion that there are always a few uncovered values of x proving that the number of primes of form x^2 + 1 are infinetely many.

Before proceeding with some demonstrative computations let me emphasise the following points:
i) It is only the non-redundant failure functions that
play a part in the computations of fractions of the interval X, X + P not covered by the preceeeding failure functions. What are redundant & non-redundant failure functions? Take the case when x =12. f(x) = 145; the relevant failure functions are 12 + 5k and 12 + 29k. 12 + 5k is redundant and 12 + 29k is non-redundant.

ii) Since the first failure function, 1 + 2k, covers all
odd values of x, only even values of x are considered in computing the residual fraction of P.
Demonstration of computation of residual fractions of the interval X, X + P, not covered by the preceeding failure functions:
The second f.f. is 2 + 5k i.e. one-fifth of the even values of P or P/10 plays a part. In other words the residual fraction is (P/2 -P/10) or 2P/5 or 0.4 P. At the next stage we get 3+ 2k & 3 + 5k. Hence the residual fraction is 0.4P - 0.4(1/10) = 0.36P. Here note that i) only the even values of x are considered and ii) only the non-redundant failure function is considered.

Before concluding let me emphasise :
i) Since atleast one value of x is not covered by the preceeding failure functions, the new value of P, in each iteration is bigger than the preceeding one.
ii) The residual fraction not covered by the preceeding failure functions diminishes at an increasingly slow rate since a bigger P implies a smaller 1/P.
iii) In view of the above two points we can safely conclude that there is always a small fraction of the interval X, X + P not covered by the preceeding failure functions.
iv) Case I and case II are mutually exclusive. Case I cannot occur as long as case II is operative.

I think this proves that there are infinitely many primes of the form x^2 + 1.

To show that the iteration is perpetual I propose to
indicate below the manner in which the fraction of the ever-growing P gets decreasing at an increasingly slower rate leaving a considerable number of values of x skiped by the failure functions.
0.36P*(33/34) = 0.3494P
0.3494P*(25/26) = 0.3360P
0.3360P*(73/74) = 0.3315P
0.3315P*(25/26) = 0.3188P
0.3188P*(81/82) = 0.3149P
0.3149P*(201/202) = 0.3133P
0.3133P*(121/122) = 0.3107P

Ultimately the fraction becomes asymptotic to a rational number significantly greater than zero. In view of the ever growing size of P this can only mean the skiping of some values of x by the failure functions which implies that a) the iteration is perpetual and b) there are infinitely many primes of the form x^2 + 1.

(concluded).

Further comments:

a) Excepting the first failure function, 1 + 2k, all the rest occur in pairs of parallel or non-intersecting failure functions.
Examples: 2 + 5k, 3 + 5k
4 + 17k, 13 + 17k etc.
In other words each pair of parallel f.f.s generate two
disjoint sets of values of x, which when substituted in f(x), yield failures ( composites).

b) Consequently we get the following shape for the rate of depletion of P ( which is ever-increasing ):

p/2( 1-1/5)^2*(1-1/13)^2*(1-1/17)^2....

This ensures that at no stage the discrete interval
X, X + P HAVE ALL VALUES OF X in that interval covered
by the prior failure functions. Since there is atleast
one value of x not covered by the f.f.s we get a new larger value of X and the corresponding P. This fuels the perpetual iteration.

Looks like no one has read the draft proof!

> To show that the iteration is perpetual I propose to
> indicate below the manner in which the fraction of
> the ever-growing P gets decreasing at an increasingly
> slower rate leaving a considerable number of values
> of x skiped by the failure functions.
> 0.36P*(33/34) = 0.3494P
> 0.3494P*(25/26) = 0.3360P
> 0.3360P*(73/74) = 0.3315P
> 0.3315P*(25/26) = 0.3188P
> 0.3188P*(81/82) = 0.3149P
> 0.3149P*(201/202) = 0.3133P
> 0.3133P*(121/122) = 0.3107P
>
> Ultimately the fraction becomes asymptotic to a
> rational number significantly greater than zero. In view
> of the ever growing size of P this can only mean the
> skiping of some values of x by the failure functions
> which implies that a) the iteration is perpetual and
> b) there are infinitely many primes of the form x^2 +
> 1.
>
> (concluded).

Let this rational number be termed "alpha".

> To show that the iteration is perpetual I propose to
> indicate below the manner in which the fraction of
> the ever-growing P gets decreasing at an increasingly
> slower rate leaving a considerable number of values
> of x skiped by the failure functions.
> 0.36P*(33/34) = 0.3494P
> 0.3494P*(25/26) = 0.3360P
> 0.3360P*(73/74) = 0.3315P
> 0.3315P*(25/26) = 0.3188P
> 0.3188P*(81/82) = 0.3149P
> 0.3149P*(201/202) = 0.3133P
> 0.3133P*(121/122) = 0.3107P
>
> Ultimately the fraction becomes asymptotic to a
> rational number significantly greater than zero. In view
> of the ever growing size of P this can only mean the
> skiping of some values of x by the failure functions
> which implies that a) the iteration is perpetual and
> b) there are infinitely many primes of the form x^2 +
> 1.
>
> (concluded).

Let this rational number be termed "alpha".

Since the function is a polynomial only failure functions
pertaining to polynomials are relevant:

x...........f(x)...............failure functions
1............2....................1+2*k -
2.............5.....................2 + 5*k

3.............10....................3+2*k,3+5*k

Note i) k belongs to Z. ii)When f(x) is composite, each factor
contributes a f.f.

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