The predicate of the current mathematical knowledge increases the constructive mathematics what is impossible for the empirical sciences

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The predicate of the current mathematical knowledge increases the constructive mathematics what is impossible for the empirical sciences

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A peer-reviewed article of this preprint also exists.

Apoloniusz Tyszka^*

This version is not peer-reviewed

Submitted:

27 July 2023

Posted:

31 July 2023

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Abstract

This is an expanded and revised version of the article: A. Tyszka, Statements and open problems on decidable sets X⊆N, Pi Mu Epsilon J. 15 (2023), no. 8, 493-504. The main results were presented at the 25th Conference Applications of Logic in Philosophy and the Foundations of Mathematics, see http://applications-of-logic.uni.wroc.pl/XXV-Konferencja-Zastosowania-Logiki-w-Filozofii-i-Podstawach-Matematyki. We assume that the current mathematical knowledge is a finite set of statements which is time-dependent. Nicolas D. Goodman observed that epistemic notions increase the understanding of mathematics without changing its content. We explain the distinction between algorithms whose existence is provable in ZFC and constructively defined algorithms which are currently known. By using this distinction, we obtain non-trivial statements on decidable sets X⊆N that belong to constructive mathematics and refer to the current mathematical knowledge on X. This and the next sentence justify the article title. For any empirical science, we can identify the current knowledge with that science because truths from the empirical sciences are not necessary truths but working models of truth from a particular context. Edmund Landau's conjecture states that the set P(n^2+1) of primes of the form n^2+1 is infinite. Landau's conjecture implies the following unproven statement Φ: card(P(n^2+1))<ω ⇒ P(n^2+1)⊆[2,(((24!)!)!)!]. We heuristically justify the statement Φ. This justification does not yield the finiteness/infiniteness of P(n^2+1). We present a new heuristic argument for the infiniteness of P(n^2+1), which is not based on the statement Φ.

Keywords:

Subject: Computer Science and Mathematics - Logic

1. Introduction

This is an expanded and revised version of the article [18]. The main results of this article were presented at the 25th Conference Applications of Logic in Philosophy and the Foundations of Mathematics, see http://applications-of-logic.uni.wroc.pl/XXV-Konferencja-Zastosowania-Logiki-w-Filozofii-i-Podstawach-Matematyki. We assume that the current mathematical knowledge is a finite set of statements which is time-dependent. Nicolas D. Goodman in [5] observed that epistemic notions increase the understanding of mathematics without changing its content. We present non-trivial statements on decidable sets

X \subseteq N

that belong to constructive mathematics and refer to the current mathematical knowledge on

X

. This and the next sentence justify the article title. For any empirical science, we can identify the current knowledge with that science because truths from the empirical sciences are not necessary truths but working models of truth from a particular context.

2. Basic definitions

Algorithms always terminate. Semi-algorithms may not terminate. There is the distinction between existing algorithms (i.e. algorithms whose existence is provable in

Z F C

) and known algorithms (i.e. algorithms whose definition is constructive and currently known), see [2,12], [14]. A definition of an integer n is called constructive, if it provides a known algorithm with no input that returns n. Definition 1 applies to sets

X \subseteq N

whose infiniteness is false or unproven.

Definition 1.

We say that a non-negative integer k is a known element of

X

, if

k \in X

and we know an algebraic expression that defines k and consists of the following signs: 1 (one), + (addition), − (subtraction), · (multiplication), ^ (exponentiation with exponent in

N

), ! (factorial of a non-negative integer), ( (left parenthesis), ) (right parenthesis).

The set of known elements of

X

is finite and time-dependent, so cannot be defined in the formal language of classical mathematics. Let t denote the largest twin prime that is smaller than ((((((((9!)!)!)!)!)!)!)!)!. The number t is an unknown element of the set of twin primes.

Definition 2.

Conditions(1)-(5)concern sets

X \subseteq N

(1)A known algorithm with no input returns an integer n satisfying

card (X) < ω \Rightarrow

X \subseteq (- \infty, n]

(2)A known algorithm for every

k \in N

decides whether or not

k \in X

(3)No known algorithm with no input returns the logical value of the statement

card (X) = ω

(4)There are many elements of

X

and it is conjectured, though so far unproven, that

X

is infinite.

(5)

X

is naturally defined. The infiniteness of

X

is false or unproven.

X

has the simplest definition among known sets

Y \subseteq N

with the same set of known elements.

Condition (3) implies that no known proof shows the finiteness/infiniteness of

X

. No known set

X \subseteq N

satisfies Conditions (1)-(4) and is widely known in number theory or naturally defined, where this term has only informal meaning.

Example 1.

The set

X = P_{n^{2} + 1}

satisfies Condition (3) .

Let

[\cdot]

denote the integer part function.

Example 2.

The set

X = \{\begin{matrix} N, & if [\frac{((((((((9!)!)!)!)!)!)!)!)!}{π}] i s o d d \\ \emptyset, & o t h e r w i s e \end{matrix}

does not satisfy Condition (3)because we know an algorithm with no input that computes

[\frac{((((((((9!)!)!)!)!)!)!)!)!}{π}]

. The set of known elements of

X

is empty. Hence, Condition(5)fails for

X

Example 3.([2,12], [14]). The function

N ∋ n \overset{h}{⟶} \{\begin{matrix} 1, & i f t h e d e c i m a l e x p a n s i o n o f π c o n t a i n s n c o n s e c u t i v e z e r o s \\ 0, & o t h e r w i s e \end{matrix}

is computable because

h = N \times {1}

or there exists

k \in N

such that

h = ({0, \dots, k} \times {1}) \cup ({k + 1, k + 2, k + 3, \dots} \times {0})

No known algorithm computes the function h.

Example 4.

The set

X = \{\begin{matrix} N, & i f t h e c o n t i n u u m h y p o t h e s i s h o l d s \\ \emptyset, & o t h e r w i s e \end{matrix}

is decidable. This

X

satisfies Conditions(1)and(3)and does not satisfy Conditions(2),(4), and(5). These facts will hold forever.

Statement 1. Condition (1) remains unproven for

X = P_{n^{2} + 1}

Proof.

For every set

X \subseteq N

, there exists an algorithm

Alg (X)

with no input that returns

n = \{\begin{matrix} 0, & if card (X) \in {0, ω} \\ max (X), & otherwise \end{matrix}

This n satisfies the implication in Condition (1), but the algorithm

Alg (P_{n^{2} + 1})

is unknown because its definition is ineffective. □

3. Main results

Edmund Landau’s conjecture states that the set

P_{n^{2} + 1}

of primes of the form

n^{2} + 1

is infinite, see [15,16,20].

Statement 2. The statement

\exists n \in N (card (P_{n^{2} + 1}) < ω \Rightarrow P_{n^{2} + 1} \subseteq [2, n + 3])

remains unproven in

Z F C

and classical logic without the law of excluded middle.

Let

f (1) = 10^{6}

, and let

f (n + 1) = f {(n)}^{f (n)}

for every positive integer n.

Statement 3. The set

X = {k \in N : (10^{6} < k) \Rightarrow (f (10^{6}), f (k)) \cap P_{n^{2} + 1} \neq \emptyset}

satisfies Conditions (1)-(4). Condition (5) fails for

X

Proof.

Condition (4) holds as

X \supseteq {0, \dots, 10^{6}}

and the set

P_{n^{2} + 1}

is conjecturally infinite. Due to known physics we are not able to confirm by a direct computation that some element of

P_{n^{2} + 1}

is greater than

f (10^{6})

, see [9]. Thus Condition (3) holds. Condition (2) holds trivially. Since the set

{k \in N : (10^{6} < k) \land (f (10^{6}), f (k)) \cap P_{n^{2} + 1} \neq \emptyset}

is empty or infinite, Condition (1) holds with

n = 10^{6}

. Condition (5) fails as the set of known elements of

X

equals

{0, \dots, 10^{6}}

. □

Statements 4 and 6 provide stronger examples.

Conjecture 1. ([1,6]). The are infinitely many primes of the form

k! + 1

For a non-negative integer n, let

ρ (n)

denote

29.5 + \frac{11!}{3 n + 1} \cdot sin (n)

Statement 4. The set

X = {n \in N : t h e i n t e r v a l [- 1, n] c o n t a i n s m o r e t h a n ρ (n) p r i m e s o f t h e f o r m k! + 1}

satisfies Conditions (1)-(5) except the requirement that

X

is naturally defined.

501893 \in X

. Condition (1) holds with

n = 501893

card (X \cap [0, 501893]) = 159827

X \cap [501894, \infty) =

{n \in N :

t h e

i n t e r v a l

[- 1, n]

c o n t a i n s

a t

l e a s t

p r i m e s

o f

t h e

f o r m

k! + 1}

Proof.

For every integer

n ⩾ 11!

, 30 is the smallest integer greater than

ρ (n)

. By this, if

n \in X \cap [11!, \infty)

, then

n + 1, n + 2, n + 3, \dots \in X

. Hence, Condition (1) holds with

n = 11! - 1

. We explicitly know 24 positive integers k such that

k! + 1

is prime, see [4]. The inequality

card ({k \in N ∖ {0} : k! + 1 i s p r i m e}) > 24

remains unproven. Since

24 < 30

, Condition (3) holds. The interval

[- 1, 11! - 1]

contains exactly three primes of the form

k! + 1

1! + 1

2! + 1

3! + 1

. For every integer

n > 503000

, the inequality

ρ (n) > 3

holds. Therefore, the execution of the following MuPAD code

m:=0:

for n from 0.0 to 503000.0 do

if n<1!+1 then r:=0 end_if:

if n>=1!+1 and n<2!+1 then r:=1 end_if:

if n>=2!+1 and n<3!+1 then r:=2 end_if:

if n>=3!+1 then r:=3 end_if:

if r>29.5+(11!/(3*n+1))*sin(n) then

m:=m+1:

print([n,m]):

end_if:

end_for:

displays the all known elements of

X

. The output ends with the line

[501893.0, 159827]

, which proves Condition (1) with

n = 501893

and Condition (4) with

card (X) ⩾ 159827

. □

Definition 3.

Conditions(1a)-(5a)concern sets

X \subseteq N

(1a)A known algorithm with no input returns an integer n satisfying

card (X) < ω \Rightarrow

X \subseteq (- \infty, n]

(2a)A known algorithm for every

k \in N

decides whether or not

k \in X

(3a)No known algorithm with no input returns the logical value of the statement

card (X) < ω

(4a)There are many elements of

X

and it is conjectured, though so far unproven, that

X

is finite.

(5a)

X

is naturally defined. The finiteness of

X

is false or unproven.

X

has the simplest definition among known sets

Y \subseteq N

with the same set of known elements.

Statement 5. The set

X = {n \in N : t h e i n t e r v a l [- 1, n] c o n t a i n s m o r e t h a n

6.5 + \frac{10^{6}}{3 n + 1} \cdot sin (n) s q u a r e s o f t h e f o r m k! + 1}

satisfies Conditions (1a)-(5a) except the requirement that

X

is naturally defined.

95151 \in X

. Condition (1a) holds with

n = 95151

card (X \cap [0, 95151]) = 30311

X \cap [95152, \infty) =

{n \in N :

t h e

i n t e r v a l

[- 1, n]

c o n t a i n s

a t

l e a s t

s q u a r e s

o f

t h e

f o r m

k! + 1}

Proof.

For every integer

n > 10^{6}

, 7 is the smallest integer greater than

6.5 + \frac{10^{6}}{3 n + 1} \cdot sin (n)

. By this, if

n \in X \cap (10^{6}, \infty)

, then

n + 1, n + 2, n + 3, \dots \in X

. Hence, Condition (1a) holds with

n = 10^{6}

. It is conjectured that

k! + 1

is a square only for

k \in {4, 5, 7}

, see [19]. Hence, the inequality

card ({k \in N ∖ {0} : k! + 1 i s a s q u a r e}) > 3

remains unproven. Since

3 < 7

, Condition (3a) holds. The interval

[- 1, 10^{6}]

contains exactly three squares of the form

k! + 1

4! + 1

5! + 1

7! + 1

. Therefore, the execution of the following MuPAD code

m:=0:

for n from 0.0 to 1000000.0 do

if n<25 then r:=0 end_if:

if n>=25 and n<121 then r:=1 end_if:

if n>=121 and n<5041 then r:=2 end_if:

if n>=5041 then r:=3 end_if:

if r>6.5+(1000000/(3*n+1))*sin(n) then

m:=m+1:

print([n,m]):

end_if:

end_for:

displays the all known elements of

X

. The output ends with the line

[95151.0, 30311]

, which proves Condition (1a) with

n = 95151

and Condition (4a) with

card (X) ⩾ 30311

. □

To formulate Statement Section 3 and its proof, we need some lemmas. For a non-negative integer n, let

θ (n)

denote the largest integer divisor of

10^{10^{10}}

smaller than n. For a non-negative integer n, let

θ_{1} (n)

denote the largest integer divisor of

10^{10}

smaller than n.

Lemma 1.

For every integer

j > 10^{10^{10}}

θ (j) = 10^{10^{10}}

. For every integer

j > 10^{10}

θ_{1} (j) = 10^{10}

Lemma 2.

For every integer

j \in (6553600, 7812500]

θ (j) = 6553600

Proof.

6553600 equals

2^{18} \cdot 5^{2}

and divides

10^{10^{10}}

7812500 < 2^{24}

7812500 < 5^{10}

. We need to prove that every integer

j \in (6553600, 7812500)

does not divide

10^{10^{10}}

. It holds as the set

\{2^{u} \cdot 5^{v} : (u \in {0, \dots, 23}) \land (v \in {0, \dots, 9})\}

contains 6553600 and 7812500 as consecutive elements. □

Lemma 3.

The number

6553600^{2} + 1

is prime.

Proof.

The following PARI/GP ([10]) command

isprime(6553600^2+1,{flag=2})

returns 1. This command performs the APRCL primality test, the best deterministic primality test algorithm ([21]). It rigorously shows that the number

6553600^{2} + 1

is prime. □

In the next lemmas, the execution of the command isprime(n,{flag=2}) proves the primality of n. Let

κ

denote the function

N ∋ n \overset{κ}{⟶} t h e_e x p o n e n t_o f_2_i n_t h e_p r i m e_f a c t o r i z a t i o n_o f_\underset{︸}{n + 1} \in N

Lemma 4.

The set

X_{1} = {n \in N : {(θ_{1} (n) + κ (n))}^{2} + 1 i s p r i m e}

is infinite.

Proof.

Let

i = 142101504

. By the inequality

2^{i} ⩾ 2 + 10^{10}

and Lemma 1, for every non-negative integer m, the number

{(θ_{1} (2^{i} \cdot (2 m + 1) - 1) + κ (2^{i} \cdot (2 m + 1) - 1))}^{2} + 1 = {(10^{10} + i)}^{2} + 1

is prime. □

Before Open Problem Section 3,

X

denotes the set

{n \in N : {(θ (n) + κ (n))}^{2} + 1 i s p r i m e}

Lemma 5.

For every

n \in X \cap (10^{10^{10}}, \infty)

and for every non-negative integer j,

3^{j} \cdot (n + 1) - 1 \in X \cap (10^{10^{10}}, \infty)

Proof.

By the inequality

3^{j} \cdot (n + 1) - 1 ⩾ n

and Lemma 1,

θ (3^{j} \cdot (n + 1) - 1) + κ (3^{j} \cdot (n + 1) - 1) = 10^{10^{10}} + κ (n) = θ (n) + κ (n)

□

Lemma 6.

card (X) ⩾ 629450

Proof.

By Lemmas 2 and 3, for every even integer

j \in (6553600, 7812500]

, the number

{(θ (j) + κ (j))}^{2} + 1 = {(6553600 + 0)}^{2} + 1

is prime. Hence,

{2 k : k \in N} \cap (6553600, 7812500] \subseteq X

Consequently,

card (X) ⩾ card ({2 k : k \in N} \cap (6553600, 7812500]) = \frac{7812500 - 6553600}{2} = 629450

□

Lemma 7.

10242 \in X

and

10242 \notin X_{1}

Proof.

The number

10240 = 2^{11} \cdot 5

divides

10^{10^{10}}

. Hence,

θ (10242) = 10240

. The number

{(θ (10242) + κ (10242))}^{2} + 1 = {(10240 + 0)}^{2} + 1

is prime. The set

\{2^{u} \cdot 5^{v} : (u \in {0, \dots, 10}) \land (v \in {0, \dots, 10})\}

contains 10000 and 12500 as consecutive elements. Hence,

θ_{1} (10242) = 10000

. The number

{(θ_{1} (10242) + κ (10242))}^{2} + 1 = {(10000 + 0)}^{2} + 1 = 17 \cdot 5882353

is composite. □

The set

X

satisfies Conditions (1)-(5) except the requirement that

X

is naturally defined.

Proof.

Condition (2) holds trivially. Let

δ

denote

10^{10^{10}}

. By Lemma 5, Condition (1) holds for

n = δ

. Lemma 5 and the unproven statement

P_{n^{2} + 1} \cap [δ^{2} + 1, \infty) \neq \emptyset

show Condition (3). The same argument and Lemma 6 yield Condition (4). By Lemma 4, the set

X_{1}

is infinite. Since Definition 1 applies to sets

X \subseteq N

whose infiniteness is false or unproven, Condition (5) holds except the requirement that

X

is naturally defined. □

The set

X

satisfies Condition (5) except the requirement that

X

is naturally defined. It is true because

X_{1}

is infinite by Lemma 4 and Definition 1 applies only to sets

X \subseteq N

whose infiniteness is false or unproven. Ignoring this restriction,

X

still satisfies the same identical condition due to Lemma 7.

Proposition 1.

No set

X \subseteq N

will satisfy Conditions(1)-(4)forever, if for every algorithm with no input, at some future day, a computer will be able to execute this algorithm in 1 second or less.

Proof.

The proof goes by contradiction. We fix an integer n that satisfies Condition (1). Since Conditions (1)-(3) will hold forever, the semi-algorithm in Figure 1 never terminates and sequentially prints the following sentences:

(T) n + 1 \notin X, n + 2 \notin X, n + 3 \notin X, \dots

(1)

The sentences from the sequence (T) and our assumption imply that for every integer

m > n

computed by a known algorithm, at some future day, a computer will be able to confirm in 1 second or less that

(n, m] \cap X = \emptyset

. Thus, at some future day, numerical evidence will support the conjecture that the set

X

is finite, contrary to the conjecture in Condition (4). □

The physical limits of computation ([9]) disprove the assumption of Proposition 1.

Open Problem 1. Is there a set

X \subseteq N

which satisfies Conditions (1)-(5)?

Open Problem 1 asks about the existence of a year

t ⩾ 2023

in which the conjunction

(C o n d i t i o n 1) \land (C o n d i t i o n 2) \land (C o n d i t i o n 3) \land (C o n d i t i o n 4) \land (C o n d i t i o n 5)

will hold for some

X \subseteq N

. For every year

t ⩾ 2023

and for every

i \in {1, 2, 3}

, a positive solution to Open Problem i in the year t may change in the future. Currently, the answers to Open Problems 1–5 are negative.

4. Satisfiable conjunctions which consist of Conditions `(1)-(5)` and their negations

The set

X = P_{n^{2} + 1}

satisfies the conjunction

\neg (C o n d i t i o n 1) \land (C o n d i t i o n 2) \land (C o n d i t i o n 3) \land (C o n d i t i o n 4) \land (C o n d i t i o n 5)

The set

X = {0, \dots, 10^{6}} \cup P_{n^{2} + 1}

satisfies the conjunction

\neg (C o n d i t i o n 1) \land (C o n d i t i o n 2) \land (C o n d i t i o n 3) \land (C o n d i t i o n 4) \land \neg (C o n d i t i o n 5)

The numbers

2^{2^{k}} + 1

are prime for

k \in {0, 1, 2, 3, 4}

. It is open whether or not there are infinitely many primes of the form

2^{2^{k}} + 1

, see [8] and [13]. It is open whether or not there are infinitely many composite numbers of the form

2^{2^{k}} + 1

, see [8] and [13]. Most mathematicians believe that

2^{2^{k}} + 1

is composite for every integer

k ⩾ 5

, see [7].

The set

X = \{\begin{matrix} N, i f 2^{2^{f (9^{9})}} + 1 i s c o m p o s i t e \\ {0, \dots, 10^{6}}, o t h e r w i s e \end{matrix}

satisfies the conjunction

(C o n d i t i o n 1) \land (C o n d i t i o n 2) \land \neg (C o n d i t i o n 3) \land (C o n d i t i o n 4) \land \neg (C o n d i t i o n 5)

Open Problem 2. Is there a set

X \subseteq N

that satisfies the conjunction

(C o n d i t i o n 1) \land (C o n d i t i o n 2) \land \neg (C o n d i t i o n 3) \land (C o n d i t i o n 4) \land (C o n d i t i o n 5) ?

The set

X = \{\begin{matrix} N, i f 2^{2^{f (9^{9})}} + 1 i s c o m p o s i t e \\ {0, \dots, 10^{6}} \cup \\ {n \in N : n i s t h e s i x t h p r i m e n u m b e r o f t h e f o r m 2^{2^{k}} + 1}, o t h e r w i s e \end{matrix}

satisfies the conjunction

\neg (C o n d i t i o n 1) \land (C o n d i t i o n 2) \land \neg (C o n d i t i o n 3) \land (C o n d i t i o n 4) \land \neg (C o n d i t i o n 5)

Open Problem 3. Is there a set

X \subseteq N

that satisfies the conjunction

\neg (C o n d i t i o n 1) \land (C o n d i t i o n 2) \land \neg (C o n d i t i o n 3) \land (C o n d i t i o n 4) \land (C o n d i t i o n 5) ?

It is possible, although very doubtful, that at some future day, the set

X = P_{n^{2} + 1}

will solve Open Problem 2. The same is true for Open Problem 3. It is possible, although very doubtful, that at some future day, the set

X = {k \in N : 2^{2^{k}} + 1 i s c o m p o s i t e}

will solve Open Problem 1. The same is true for Open Problems 2 and 3.

Table 1 shows satisfiable conjunctions of the form

# (Condition 1) \land (Condition 2) \land # (Condition 3) \land (Condition 4) \land # (Condition 5)

where # denotes the negation ¬ or the absence of any symbol. Table 1 differs from Table 1 in [18] for three sets

X

. These sets

X

have the index

n e w

Definition 4.

We say that an integer n is a threshold number of a set

X \subseteq N

, if

card (X) < ω \Rightarrow X \subseteq (- \infty, n]

If a set

X \subseteq N

is empty or infinite, then any integer n is a threshold number of

X

. If a set

X \subseteq N

is non-empty and finite, then the all threshold numbers of

X

form the set

[max (X), \infty) \cap N

Open Problem 4. Is there a known threshold number of

P_{n^{2} + 1}

Open Problem 4 asks about the existence of a year

t ⩾ 2023

in which the implication

card (P_{n^{2} + 1}) < ω \Rightarrow P_{n^{2} + 1} \subseteq (- \infty, n]

will hold for some known integer n.

Let

T

denote the set of twin primes. Is there a known threshold number of

T

Open Problem 5 asks about the existence of a year

t ⩾ 2023

in which the implication

card (T) < ω \Rightarrow T \subseteq (- \infty, n]

will hold for some known integer n.

5. Number-theoretic statements $Ψ_{n}$

Let

f (1) = 2

f (2) = 4

, and let

f (n + 1) = f (n)!

for every integer

n ⩾ 2

. Let

U_{1}

denote the system of equations

{x_{1}! = x_{1}

. For an integer

n ⩾ 2

, let

U_{n}

denote the following system of equations:

\{\begin{matrix} x_{1}! & = & x_{1} \\ x_{1} \cdot x_{1} & = & x_{2} \\ \forall i \in {2, \dots, n - 1} x_{i}! & = & x_{i + 1} \end{matrix}

Lemma 8.

For every positive integer n, the system

U_{n}

has exactly two solutions in positive integers

x_{1}, \dots, x_{n}

, namely

(1, \dots, 1)

and

(f (1), \dots, f (n))

Let

B_{n}

denote the following system of equations:

{x_{j}! = x_{k} : j, k \in {1, \dots, n}} \cup {x_{i} \cdot x_{j} = x_{k} : i, j, k \in {1, \dots, n}}

For every positive integer n, no known system

S \subseteq B_{n}

with a finite number of solutions in positive integers

x_{1}, \dots, x_{n}

has a solution

(x_{1}, \dots, x_{n}) \in {(N ∖ {0})}^{n}

satisfying

max (x_{1}, \dots, x_{n}) > f (n)

. For every positive integer n and for every known system

S \subseteq B_{n}

, if the finiteness/infiniteness of the set

{(x_{1}, \dots, x_{n}) \in {(N ∖ {0})}^{n} : (x_{1}, \dots, x_{n}) s o l v e s S}

is unknown, then the statement

\exists x_{1}, \dots, x_{n} \in N ∖ {0} ((x_{1}, \dots, x_{n}) s o l v e s S) \land (max (x_{1}, \dots, x_{n}) > f (n))

remains unproven.

For a positive integer n, let

Ψ_{n}

denote the following statement: if a system

S \subseteq B_{n}

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{n}

, then each such solution

(x_{1}, \dots, x_{n})

satisfies

x_{1}, \dots, x_{n} ⩽ f (n)

. The statement

Ψ_{n}

says that for subsystems of

B_{n}

with a finite number of solutions, the largest known solution is indeed the largest possible. The statement

\forall n \in N ∖ {0} Ψ_{n}

is dubious, see [17].

Theorem 1.

For every statement

Ψ_{n}

, the bound

f (n)

cannot be decreased.

Proof.

It follows from Lemma 8 because

U_{n} \subseteq B_{n}

. □

Theorem 2.

For every integer

n ⩾ 2

, the statement

Ψ_{n + 1}

implies the statement

Ψ_{n}

Proof.

If a system

S \subseteq B_{n}

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{n}

, then for every integer

i \in {1, \dots, n}

the system

S \cup {x_{i}! = x_{n + 1}}

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{n + 1}

. The statement

Ψ_{n + 1}

implies that

x_{i}! = x_{n + 1} ⩽ f (n + 1) = f (n)!

. Hence,

x_{i} ⩽ f (n)

. □

Theorem 3.

Every statement

Ψ_{n}

is true with an unknown integer bound that depends on n.

Proof.

For every positive integer n, the system

B_{n}

has a finite number of subsystems. □

6. A special case of the statement $Ψ_{9}$ applies to the conjecture that $card (P_{n^{2} + 1}) = ω$

Let

A

denote the following system of equations:

\{\begin{matrix} x_{2}! & = & x_{3} \\ x_{3}! & = & x_{4} \\ x_{5}! & = & x_{6} \\ x_{8}! & = & x_{9} \\ x_{1} \cdot x_{1} & = & x_{2} \\ x_{3} \cdot x_{5} & = & x_{6} \\ x_{4} \cdot x_{8} & = & x_{9} \\ x_{5} \cdot x_{7} & = & x_{8} \end{matrix}

Lemma 9.

For every positive integers x and y,

x! \cdot y = y!

if and only if

(x + 1 = y) \lor (x = y = 1)

Lemma 9 and the diagram in Figure 2 explain the construction of the system

A

Lemma 10.(Wilson’s theorem, [3]). For every integer

x ⩾ 2

, x is prime if and only if x divides

(x - 1)! + 1

Lemma 11.

For every integer

x_{1} ⩾ 2

, the system

A

is solvable in positive integers

x_{2}, \dots, x_{9}

if and only if

x_{1}^{2} + 1

is prime. In this case, the integers

x_{2}, \dots, x_{9}

are uniquely determined by the following equalities:

\begin{matrix} x_{2} & = & x_{1}^{2} \\ x_{3} & = & (x_{1}^{2})! \\ x_{4} & = & ((x_{1}^{2})!)! \\ x_{5} & = & x_{1}^{2} + 1 \\ x_{6} & = & (x_{1}^{2} + 1)! \\ x_{7} & = & \frac{(x_{1}^{2})! + 1}{x_{1}^{2} + 1} \\ x_{8} & = & (x_{1}^{2})! + 1 \\ x_{9} & = & ((x_{1}^{2})! + 1)! \end{matrix}

Proof.

By Lemma 9, for every integer

x_{1} ⩾ 2

, the system

A

is solvable in positive integers

x_{2}, \dots, x_{9}

if and only if

x_{1}^{2} + 1

divides

(x_{1}^{2})! + 1

. Hence, the claim of Lemma 11 follows from Lemma 10. □

Lemma 12.

There are only finitely many tuples

(x_{1}, \dots, x_{9}) \in {(N ∖ {0})}^{9}

, which solve the system

A

and satisfy

x_{1} = 1

. It is true as every such tuple

(x_{1}, \dots, x_{9})

satisfies

x_{1}, \dots, x_{9} \in {1, 2}

Proof.

The equality

x_{1} = 1

implies that

x_{2} = x_{1} \cdot x_{1} = 1

. Hence,

x_{3} = x_{2}! = 1

. Therefore,

x_{4} = x_{3}! = 1

. The equalities

x_{5}! = x_{6}

and

x_{5} = 1 \cdot x_{5} = x_{3} \cdot x_{5} = x_{6}

imply that

x_{5}, x_{6} \in {1, 2}

. The equalities

x_{8}! = x_{9}

and

x_{8} = 1 \cdot x_{8} = x_{4} \cdot x_{8} = x_{9}

imply that

x_{8}, x_{9} \in {1, 2}

. The equality

x_{5} \cdot x_{7} = x_{8}

implies that

x_{7} = \frac{x_{8}}{x_{5}} \in

\{\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{2}{2}\} \cap (N ∖ {0}) = {1, 2}

. □

Conjecture 2. The statement

Ψ_{9}

is true when is restricted to the system

A

Theorem 4.

Conjecture 2 proves the following implication: if there exists an integer

x_{1} ⩾ 2

such that

x_{1}^{2} + 1

is prime and greater than

f (7)

, then the set

P_{n^{2} + 1}

is infinite.

Proof.

Suppose that the antecedent holds. By Lemma 11, there exists a unique tuple

(x_{2}, \dots, x_{9}) \in {(N ∖ {0})}^{8}

such that the tuple

(x_{1}, x_{2}, \dots, x_{9})

solves the system

A

. Since

x_{1}^{2} + 1 > f (7)

, we obtain that

x_{1}^{2} ⩾ f (7)

. Hence,

(x_{1}^{2})! ⩾ f (7)! = f (8)

. Consequently,

x_{9} = ((x_{1}^{2})! + 1)! ⩾ (f (8) + 1)! > f (8)! = f (9)

Conjecture 2 and the inequality

x_{9} > f (9)

imply that the system

A

has infinitely many solutions (

x_{1}, \dots, x_{9} {) \in (N ∖ {0})}^{9}

. According to Lemmas 11 and 12, the set

P_{n^{2} + 1}

is infinite. □

Landau’s conjecture implies the following unproven statement

Φ

card (P_{n^{2} + 1}) < ω \Rightarrow P_{n^{2} + 1} \subseteq [2, (((24!)!)!)!]

Theorem 5 heuristically justifies the statement

Φ

. This justification does not yield the finiteness/infiniteness of

P_{n^{2} + 1}

Theorem 5.

Conjecture 2 implies the statement Φ.

Proof.

It follows from Theorem 4 and the equality

f (7) = (((24!)!)!)!

. □

Theorem 6.

The statement Φ implies Conjecture 2.

Proof.

By Lemmas 11 and 12, if positive integers

x_{1}, \dots, x_{9}

solve the system

A

, then

(x_{1} ⩾ 2) \land (x_{5} = x_{1}^{2} + 1) \land (x_{5} is prime)

x_{1}, \dots, x_{9} \in {1, 2}

. In the first case, Lemma 11 and the statement

Φ

imply that the inequality

x_{5} ⩽ (((24!)!)!)! = f (7)

holds when the system

A

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{9}

. Hence,

x_{2} = x_{5} - 1 < f (7)

and

x_{3} = x_{2}! < f (7)! = f (8)

. Continuing this reasoning in the same manner, we can show that every

x_{i}

does not exceed

f (9)

. □

Lemma 13.

{log}_{2} ({log}_{2} ({log}_{2} ({log}_{2} ({log}_{2} ({log}_{2} ({log}_{2} ((((24!)!)!)!) \approx 1.42298

Proof.

We ask Wolfram Alpha at http://wolframalpha.com. □

Statement 7. Conditions (2)–(5) hold for

X = P_{n^{2} + 1}

. The statement

Φ

implies Condition (1) for

X = P_{n^{2} + 1}

and does not falsify Conditions (2)–(5).

Proof.

Conditions (2), (3), and (5) hold trivially. The set

P_{n^{2} + 1}

is conjecturally infinite. There are 2199894223892 primes of the form

n^{2} + 1

in the interval

[2, 10^{28})

, see [16]. These two facts imply Condition (4). The statement

Φ

implies that Condition (1) holds for

X = P_{n^{2} + 1}

with

n = (((24!)!)!)!

. By Lemma 13, due to known physics we are not able to confirm by a direct computation that some element of

P_{n^{2} + 1}

is greater than

f (7) = (((24!)!)!)!

, see [9]. Hence, the statement

Φ

does not falsify Conditions (2)–(5). □

Proving Landau’s conjecture will disprove Statement Section 6. We do not conjecture that

(Conditions (1)–(5) hold for

X = P_{n^{2} + 1}) \land Φ

7. A new heuristic argument for the infiniteness of $P_{n^{2} + 1}$

The system

A

contains four factorials and four multiplications. Let

F

denote the family of all systems

S \subseteq B_{9}

which contain at most four factorials and at most four multiplications.

Among known systems

S \in F

, the following system

C

\{\begin{matrix} x_{1}! & = & x_{2} \\ x_{2} \cdot x_{9} & = & x_{1} \\ x_{2} \cdot x_{2} & = & x_{3} \\ x_{3} \cdot x_{3} & = & x_{4} \\ x_{4} \cdot x_{4} & = & x_{5} \\ x_{5}! & = & x_{6} \\ x_{6}! & = & x_{7} \\ x_{7}! & = & x_{8} \end{matrix}

attains the greatest solution in positive integers

x_{1}, \dots, x_{9}

and has at most finitely many solutions in

{(N ∖ {0})}^{9}

. Only the tuples

(1, \dots, 1)

and

(2, 2, 4, 16, 256, 256!, (256!)!, ((256!)!)!, 1)

solve

C

and belong to

{(N ∖ {0})}^{9}

For every known system

S \in F

, if the finiteness of the set

{(x_{1}, \dots, x_{9}) \in {(N ∖ {0})}^{9} : (x_{1}, \dots, x_{9}) s o l v e s S}

is unproven and conjectured, then the statement

\exists x_{1}, \dots, x_{9} \in N ∖ {0} ((x_{1}, \dots, x_{9}) s o l v e s S) \land (max (x_{1}, \dots, x_{9}) > ((256!)!)!)

remains unproven.

Let

Γ

denote the statement: if the system

A

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{9}

, then each such solution

(x_{1}, \dots, x_{9})

satisfies

x_{1}, \dots, x_{9} ⩽ ((256!)!)!

. The number

46^{512} + 1

is prime ([11]) and greater than

256!

, see also [13] for the primality of

150^{2048} + 1

. Hence, the statement

Γ

is equivalent to the infiniteness of

P_{n^{2} + 1}

. It heuristically justifies the infiniteness of

P_{n^{2} + 1}

in a sophisticated way.

References

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Factorial prime, http://en.wikipedia.org/wiki/Factorial_prime.
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Figure 1. Semi-algorithm that terminates if and only if

X

is infinite.

Figure 1. Semi-algorithm that terminates if and only if

X

is infinite.

Figure 2. Construction of the system

A

Figure 2. Construction of the system

A

Table 1. Five satisfiable conjunctions.

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The predicate of the current mathematical knowledge increases the constructive mathematics what is impossible for the empirical sciences

Abstract

1. Introduction

2. Basic definitions

3. Main results

4. Satisfiable conjunctions which consist of Conditions (1)-(5) and their negations

5. Number-theoretic statements Ψ n

6. A special case of the statement Ψ 9 applies to the conjecture that card ( P n 2 + 1 ) = ω

7. A new heuristic argument for the infiniteness of P n 2 + 1

References

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4. Satisfiable conjunctions which consist of Conditions `(1)-(5)` and their negations

5. Number-theoretic statements $Ψ_{n}$

6. A special case of the statement $Ψ_{9}$ applies to the conjecture that $card (P_{n^{2} + 1}) = ω$

7. A new heuristic argument for the infiniteness of $P_{n^{2} + 1}$