A Common Approach to Three Open Problems in Number Theory

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A Common Approach to Three Open Problems in Number Theory

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Apoloniusz Tyszka^*

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Submitted:

18 April 2023

Posted:

19 April 2023

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Abstract

The following system of equations {x_1 \cdot x_1=x_2, x_2 \cdot x_2=x_3, 2^{2^{x_1}}=x_3, x_4 \cdot x_5=x_2, x_6 \cdot x_7=x_2} has exactly one solution in ({\mathbb N}\{0,1})^7, namely (2,4,16,2,2,2,2). Hypothesis 1 states that if a system of equations S \subseteq {x_i \cdot x_j=x_k: i,j,k \in {1,...,7}} \cup {2^{2^{x_j}}=x_k: j,k \in {1,...,7}} has at most five equations and at most finitely many solutions in ({\mathbb N}\{0,1})^7, then each such solution (x_1,...,x_7) satisfies x_1,...,x_7 \leq 16. Hypothesis 1 implies that there are infinitely many composite numbers of the form 2^{2^{n}}+1. Hypotheses 2 and 3 are of similar kind. Hypothesis 2 implies that if the equation x!+1=y^2 has at most finitely many solutions in positive integers x and y, then each such solution (x,y) belongs to the set {(4,5),(5,11),(7,71)}. Hypothesis 3 implies that if the equation x(x+1)=y! has at most finitely many solutions in positive integers x and y, then each such solution (x,y) belongs to the set {(1,2),(2,3)}. We describe semi-algorithms sem_j (j=1,2,3) such that Hypothesis j holds if and only if sem_j prints consecutive positive integers starting from 1.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Composite Numbers of the Form $2^{2^{n}} + 1$

Let

A

denote the following system of equations:

\{x_{i} \cdot x_{j} = x_{k} : i, j, k \in {1, \dots, 7}\} \cup \{2^{2^{x_{j}}} = x_{k} : j, k \in {1, \dots, 7}\}

The following subsystem of

A

has exactly one solution in

{(N ∖ {0, 1})}^{7}

, namely

(2, 4, 16, 2, 2, 2, 2)

Hypothesis 1.

If a system of equations

S \subseteq A

has at most five equations and at most finitely many solutions in

{(N ∖ {0, 1})}^{7}

, then each such solution

(x_{1}, \dots, x_{7})

satisfies

x_{1}, \dots, x_{7} ⩽ 16

Lemma 1. ([7] (p. 109) ). For every non-negative integers x and y,

x + 1 = y

if and only if

2^{2^{x}} \cdot 2^{2^{x}} = 2^{2^{y}}

Theorem 1.

Hypothesis 1 implies that

2^{2^{x_{1}}} + 1

is composite for infinitely many integers

x_{1}

greater than 1.

Proof.

Assume, on the contrary, that Hypothesis 1 holds and

2^{2^{x_{1}}} + 1

is composite for at most finitely many integers

x_{1}

greater than 1. Then, the equation

x_{2} \cdot x_{3} = 2^{2^{x_{1}}} + 1

has at most finitely many solutions in

{(N ∖ {0, 1})}^{3}

. By Lemma 1, in positive integers greater than 1, the following subsystem of

A

has at most finitely many solutions in

{(N ∖ {0, 1})}^{7}

and expresses that

\{\begin{matrix} x_{2} \cdot x_{3} & = & 2^{2^{x_{1}}} + 1 \\ x_{4} & = & 2^{2^{x_{1}}} + 1 \\ x_{5} & = & 2^{2^{x_{1}}} \\ x_{6} & = & 2^{2^{2^{2^{x_{1}}}}} \\ x_{7} & = & 2^{2^{2^{2^{x_{1}}} + 1}} \end{matrix}

Since

641 \cdot 6700417 = 2^{2^{5}} + 1 > 16

, we get a contradiction. □

Most mathematicians believe that

2^{2^{n}} + 1

is composite for every integer

n ⩾ 5

, see [2] (p. 23). Open Problem 1. ([3] (p. 159)). Are there infinitely many composite numbers of the form

2^{2^{n}} + 1

Primes of the form

2^{2^{n}} + 1

are called Fermat primes, as Fermat conjectured that every integer of the form

2^{2^{n}} + 1

is prime, see [3] (p. 1). Fermat remarked that

2^{2^{0}} + 1 = 3

2^{2^{1}} + 1 = 5

2^{2^{2}} + 1 = 17

2^{2^{3}} + 1 = 257

, and

2^{2^{4}} + 1 = 65537

are all prime, see [3] (p. 1).

Open Problem 2. ([3] (p. 158)). Are there infinitely many prime numbers of the form

2^{2^{n}} + 1

2. The Brocard-Ramanujan Equation $x! + 1 = y^{2}$

Let

B

denote the following system of equations:

{x_{i} \cdot x_{j} = x_{k} : i, j, k \in {1, \dots, 6}} \cup {x_{j}! = x_{k} : (j, k \in {1, \dots, 6}) \land (j \neq k)}

The following subsystem of

B

has exactly two solutions in positive integers, namely

(1, \dots, 1)

and

(2, 2, 4, 24, 24!, (24!)!)

Hypothesis 2.

If a system of equations

S \subseteq B

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{6}

, then each such solution

(x_{1}, \dots, x_{6})

satisfies

x_{1}, \dots, x_{6} ⩽ (24!)!

Lemma 2.

For every positive integers x and y,

x! \cdot y = y!

if and only if

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

Theorem 2.

Hypothesis 2 implies that if the equation

x_{1}! + 1 = x_{2}^{2}

has at most finitely many solutions in positive integers

x_{1}

and

x_{2}

, then each such solution

(x_{1}, x_{2})

belongs to the set

{(4, 5), (5, 11), (7, 71)}

Proof.

The following system of equations

B_{1}

is a subsystem of

B

. By Lemma 2, in positive integers, the system

B_{1}

expresses that

x_{1} = \dots = x_{6} = 1

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

If the equation

x_{1}! + 1 = x_{2}^{2}

has at most finitely many solutions in positive integers

x_{1}

and

x_{2}

, then

B_{1}

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{6}

and Hypothesis 2 implies that every tuple

(x_{1}, \dots, x_{6})

of positive integers that solves

B_{1}

satisfies

(x_{1}! + 1)! = x_{6} ⩽ (24!)!

. Hence,

x_{1} \in {1, \dots, 23}

. If

x_{1} \in {1, \dots, 23}

, then

x_{1}! + 1

is a square only for

x_{1} \in {4, 5, 7}

. □

It is conjectured that

x! + 1

is a square only for

x \in {4, 5, 7}

, see [8] (p. 297). A weak form of Szpiro’s conjecture implies that the equation

x! + 1 = y^{2}

has only finitely many solutions in positive integers, see [6].

3. Erdös’ Equation $x (x + 1) = y!$

Let

C

denote the following system of equations:

{x_{i} \cdot x_{j} = x_{k} : (i, j, k \in {1, \dots, 6}) \land (i \neq j)} \cup {x_{j}! = x_{k} : (j, k \in {1, \dots, 6}) \land (j \neq k)}

The following subsystem of

C

has exactly three solutions in positive integers, namely

(1, \dots, 1)

(1, 1, 2, 2, 2, 2)

, and

(2, 2, 3, 6, 720, 720!)

Hypothesis 3.

If a system of equations

S \subseteq C

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{6}

, then each such solution

(x_{1}, \dots, x_{6}

) satisfies

x_{1}, \dots, x_{6} ⩽ 720!

Theorem 3.

Hypothesis 3 implies that if the equation

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

has at most finitely many solutions in positive integers

x_{1}

and

x_{2}

, then each such solution

(x_{1}, x_{2})

belongs to the set

{(1, 2), (2, 3)}

Proof.

The following system of equations

C_{1}

is a subsystem of

C

. By Lemma 2, in positive integers, the system

C_{1}

expresses that

x_{1} = \dots = x_{6} = 1

\{\begin{matrix} x_{1} \cdot (x_{1} + 1) & = & x_{2}! \\ x_{3} & = & x_{1} \cdot (x_{1} + 1) \\ x_{4} & = & x_{1}! \\ x_{5} & = & x_{1} + 1 \\ x_{6} & = & (x_{1} + 1)! \end{matrix}

If the equation

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

has at most finitely many solutions in positive integers

x_{1}

and

x_{2}

, then

C_{1}

has at most finitely many solutions in positive integers

x_{1}, \dots, x_{6}

and Hypothesis 3 implies that every tuple

(x_{1}, \dots, x_{6})

of positive integers that solves

C_{1}

satisfies

x_{2}! = x_{3} ⩽ 720!

. Hence,

x_{2} \in {1, \dots, 720}

. If

x_{2} \in {1, \dots, 720}

, then

x_{2}!

is a product of two consecutive positive integers only for

x_{2} \in {2, 3}

because the following MuPAD program

for x2 from 1 to 720 do
x1:=round(sqrt(x2!+(1/4))-(1/2)):
if x1*(x1+1)=x2! then print(x2) end_if:
end_for:

returns 2 and 3. □

The question of solving the equation

x (x + 1) = y!

was posed by P. Erdös, see [1]. F. Luca proved that the

a b c

conjecture implies that the equation

x (x + 1) = y!

has only finitely many solutions in positive integers, see [4].

4. There is No Hope for a Hypothesis that is Similar to Hypothesis 2 or 3 and Holds for an Arbitrary Number of Variables

Let

f (1) = 2

f (2) = 4

, and let

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

for every integer

n ⩾ 2

. Let

W_{1}

denote the system of equations

{x_{1}! = x_{1}

. For an integer

n ⩾ 2

, let

W_{n}

denote the following system of equations:

For every positive integer n, the system

W_{n}

has exactly two solutions in positive integers

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

, namely

(1, \dots, 1)

and

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

. For a positive integer n, let

Ψ_{n}

denote the following statement: if a system of equations

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

Theorem 4.

Every factorial Diophantine equation can be algorithmically transformed into an equivalent system of equations of the forms

x_{i} \cdot x_{j} = x_{k}

and

x_{j}! = x_{k}

. It means that this system of equations satisfies a modified version of Lemma 4 in [7].

Proof.

It follows from Lemmas 2–5 in [7] and Lemma 2. □

The statement

\forall n \in N ∖ {0} Ψ_{n}

is dubious. By Theorem 4, this statement implies that there is an algorithm which takes as input a factorial Diophantine equation and returns an integer which is greater than the solutions in positive integers, if these solutions form a finite set. This conclusion is strange because properties of factorial Diophantine equations are similar to properties of exponential Diophantine equations and a computable upper bound on non-negative integer solutions does not exist for exponential Diophantine equations with a finite number of solutions, see [5].

5. There are Semi-algorithms ${s e m}_{j}$ $(j = 1, 2, 3)$ Such that Hypothesis j Holds if and Only if ${s e m}_{j}$ Prints Consecutive Positive Integers Starting from 1

k \in [10^{19}, 10^{20} - 1] \cap N

, then there are uniquely determined non-negative integers

a (0), \dots, a (19) \in {0, \dots, 9}

such that

(a (19) ⩾ 1) \land (k = a (19) \cdot 10^{19} + a (18) \cdot 10^{18} + \dots + a (1) \cdot 10^{1} + a (0) \cdot 10^{0})

Definition 4.

For an integer

k \in [10^{19}, 10^{20} - 1]

S_{k}

stands for the smallest system of equations

S

satisfying conditions(1)and(2) .

(1)If

i \in {0, 4, 8, 16}

and

a (i) \in {0, 1, 2, 3, 4}

, then the equation

x_{a (i + 1)} \cdot x_{a (i + 2)} = x_{a (i + 3)}

belongs to

S

when it belongs to

A

(2)If

i \in {0, 4, 8, 16}

and

a (i) \in {5, 6, 7, 8, 9}

, then the equation

2^{2^{x_{a (i + 1)}}} = x_{a (i + 2)}

belongs to

S

when it belongs to

A

Lemma 4.

{S_{k} : k \in [10^{19}, 10^{20} - 1] \cap N} = {S : (S \subseteq A) \land (card (S) ⩽ 5)}

Proof.

It follows from the equality

5 \cdot 4 = 20

. □

For a positive integer n, let

p_{n}

denote the n-th prime number.

Theorem 5.

Hypothesis 1 holds if and only if the following semi-algorithm prints consecutive positive integers starting from 1.

Proof.

It follows from Lemma 4. □

k \in [10^{1007}, 10^{1008} - 1] \cap N

, then there are uniquely determined non-negative integers

c (0), \dots, c (1007) \in {0, \dots, 9}

such that

(c (1007 ⩾ 1) \land (k = c (1007) \cdot 10^{1007} + c (1006) \cdot 10^{1006} + \dots + c (1) \cdot 10^{1} + c (0) \cdot 10^{0})

Definition 5.

For an integer

k \in [10^{1007}, 10^{1008} - 1]

T_{k}

stands for the smallest system of equations

T

satisfying conditions(3)and(4) .

(3)If

i \in {0, 4, 8, \dots, 1004}

and

c (i) \in {0, 1, 2, 3, 4}

, then the equation

x_{c (i + 1)} \cdot x_{c (i + 2)} = x_{c (i + 3)}

belongs to

T

when it belongs to

B

(4)If

i \in {0, 4, 8, \dots, 1004}

and

c (i) \in {5, 6, 7, 8, 9}

, then the equation

x_{c (i + 1)}! = x_{c (i + 2)}

belongs to

T

when it belongs to

B

Lemma 5.

{T_{k} : k \in [10^{1007}, 10^{1008} - 1] \cap N} = {T : T \subseteq B}

Proof.

It follows from the equality

(6^{3} + 6^{2}) \cdot 4 = 1008

. □

Definition 6.

For an integer

k \in [10^{1007}, 10^{1008} - 1]

U_{k}

stands for the smallest system of equations

U

satisfying conditions(5)and(6) .

(5)If

i \in {0, 4, 8, \dots, 1004}

and

c (i) \in {0, 1, 2, 3, 4}

, then the equation

x_{c (i + 1)} \cdot x_{c (i + 2)} = x_{c (i + 3)}

belongs to

U

when it belongs to

C

(6)If

i \in {0, 4, 8, \dots, 1004}

and

c (i) \in {5, 6, 7, 8, 9}

, then the equation

x_{c (i + 1)}! = x_{c (i + 2)}

belongs to

U

when it belongs to

C

Lemma 6.

{U_{k} : k \in [10^{1007}, 10^{1008} - 1] \cap N} = {U : U \subseteq C}

Proof.

It follows from the equality

(6^{3} + 6^{2}) \cdot 4 = 1008

. □

Theorem 6.

Hypothesis 2 holds if and only if the following semi-algorithm prints consecutive positive integers starting from 1.

Proof.

It follows from Lemma 5. □

Theorem 7.

Hypothesis 3 holds if and only if the following semi-algorithm prints consecutive positive integers starting from 1.

Proof.

It follows from Lemma 6. □

References

D. Berend and J. E. Harmse, On polynomial-factorial Diophantine equations, Trans. Amer. Math. Soc. 358 (2006), no. 4, 1741–1779.
J.-M. De Koninck and F. Luca, Analytic number theory: Exploring the anatomy of integers, American Mathematical Society, Providence, RI, 2012.
M. Křížek, F. Luca, L. Somer, 17 lectures on Fermat numbers: From number theory to geometry, Springer, New York, 2001.
F. Luca, The Diophantine equation P(x)=n! and a result of M. Overholt, Glas. Mat. Ser. III 37 (57) (2002), no. 2, 269–273.
Yu. Matiyasevich, Existence of noneffectivizable estimates in the theory of exponential Diophantine equations, J. Sov. Math. vol. 8, no. 3, 1977, 299–311. [CrossRef]
M. Overholt, The Diophantine equation n!+1=m², Bull. London Math. Soc. 25 (1993), no. 2, 104.
A. Tyszka, A hypothetical upper bound on the heights of the solutions of a Diophantine equation with a finite number of solutions, Open Comput. Sci. 8 (2018), no. 1, 109–114. [CrossRef]
E. W. Weisstein, CRC Concise Encyclopedia of Mathematics, 2nd ed., Chapman & Hall/CRC, Boca Raton, FL, 2002.

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A Common Approach to Three Open Problems in Number Theory

Abstract

1. Composite Numbers of the Form 2 2 n + 1

2. The Brocard-Ramanujan Equation x ! + 1 = y 2

3. Erdös’ Equation x ( x + 1 ) = y !

4. There is No Hope for a Hypothesis that is Similar to Hypothesis 2 or 3 and Holds for an Arbitrary Number of Variables

5. There are Semi-algorithms s e m j ( j = 1 , 2 , 3 ) Such that Hypothesis j Holds if and Only if s e m j Prints Consecutive Positive Integers Starting from 1

References

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1. Composite Numbers of the Form $2^{2^{n}} + 1$

2. The Brocard-Ramanujan Equation $x! + 1 = y^{2}$

3. Erdös’ Equation $x (x + 1) = y!$

5. There are Semi-algorithms ${s e m}_{j}$ $(j = 1, 2, 3)$ Such that Hypothesis j Holds if and Only if ${s e m}_{j}$ Prints Consecutive Positive Integers Starting from 1