In this paper, we prove Legendre’s conjecture: There is a prime number between n 2 and ( n +1) 2 for every positive integer n. We also prove three related conjectures. The method that we use is to analyze binomial coefficients. It is developed by the author from the method of analyzing binomial central coefficients, that was used by Paul Erdős in his proof of Bertrand’s postulate - Chebyshev’s theorem.
Legendre’s conjecture was proposed by Andrien-Marie Legendre (1752-1833). It states that there is a prime number between n 2 and ( n + 1 ) 2 for every positive integer n. The conjecture was one of Landau’s problems on prime numbers in 1912. Many researchers have been trying to resolve this conjecture without success, as none of the empirical, asymptotic, probabilistic, and statistical methods of proving the Legendre conjecture were considered to provide sufficient evidence. According to Wikipedia, as of 2022, the conjecture has never been proved nor disproved [
We will use the method of analyzing binomial coefficients, ( λ n n ) , to prove
the Legendre’s conjecture, where λ is an integer and λ ≥ 3 . The method is developed by the author of this paper from the method of analyzing binomial
central coefficients, ( 2 n n ) , that was used by Paul Erdős [
In Section 1, we will define the prime number factorization operator and clarify some terms and concepts. In Section 2, we will derive some lemmas. In Section 3, we will develop a theorem to be used in the proofs of the conjectures in the later sections. In Section 4, we will prove Legendre’s conjecture, and in Section 5, we will prove Oppermann’s conjecture [
Definition: Γ a ≥ p > b { ( λ n n ) } denotes the prime number factorization operator of the integer expression ( λ n n ) . It is the product of the prime numbers in the decomposition of ( λ n n ) in the range of a ≥ p > b . In this operator, p is a prime number, a and b are real numbers, and λ n ≥ a ≥ p > b ≥ 1 .
It has some properties:
It is always true that Γ a ≥ p > b { ( λ n n ) } ≥ 1 (1.1)
If there is no prime number in ( λ n n ) within the range of a ≥ p > b , then Γ a ≥ p > b { ( λ n n ) } = 1 , or vice versa, if Γ a ≥ p > b { ( λ n n ) } = 1 , then there is no prime number in ( λ n n ) within the range of a ≥ p > b . (1.2)
For example, when λ = 5 and n = 4 , Γ 16 ≥ p > 10 { ( 20 4 ) } = 13 0 ⋅ 11 0 = 1 . No prime number 13 or 11 is in ( 20 4 ) in the range of 16 ≥ p > 10 .
If there is at least one prime number in ( λ n n ) in the range of a ≥ p > b , then Γ a ≥ p > b { ( λ n n ) } > 1 , or vice versa, if Γ a ≥ p > b { ( λ n n ) } > 1 , then there is at least one prime number in ( λ n n ) within the range of a ≥ p > b . (1.3)
For example, when λ = 5 and n = 4 , Γ 18 ≥ p > 16 { ( 20 4 ) } = 17 > 1 . A prime number 17 is in ( 20 4 ) within the range of 18 ≥ p > 16 .
Let v p ( n ) be the p-adic valuation of n, the exponent of the highest power of p that divides n.
Similar to Paul Erdős’ paper [
v p ( ( λ n n ) ) = v p ( ( λ n ) ! ) − v p ( ( ( λ − 1 ) n ) ! ) − v p ( n ! ) = ∑ i = 1 R ( p ) ( ⌊ λ n p i ⌋ − ⌊ ( λ − 1 ) n p i ⌋ − ⌊ n p i ⌋ ) ≤ R ( p )
because for any real numbers a and b , the expression of ⌊ a + b ⌋ − ⌊ a ⌋ − ⌊ b ⌋ is 0 or 1.
Thus, if p divides ( λ n n ) , then v p ( ( λ n n ) ) ≤ R ( p ) ≤ log p ( λ n ) , or
p v p ( ( λ n n ) ) ≤ p R ( p ) ≤ λ n (1.4)
If λ n ≥ p > ⌊ λ n ⌋ , then 0 ≤ v p ( ( λ n n ) ) ≤ R ( p ) ≤ 1 . (1.5)
Let π ( n ) be the number of distinct prime numbers less than or equal to n. Among the first six consecutive natural numbers are three prime numbers 2, 3 and 5. Then, for each additional six consecutive natural numbers, at most one can add two prime numbers, p ≡ 1 ( MOD 6 ) and p ≡ 5 ( MOD 6 ) . Thus,
π ( n ) ≤ ⌊ n 3 ⌋ + 2 ≤ n 3 + 2 . (1.6)
From the prime number decomposition, when n > ⌊ λ n ⌋ ,
( λ n n ) = Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ n ≥ p > ⌊ λ n ⌋ { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! }
when n ≤ ⌊ λ n ⌋ , ( λ n n ) ≤ Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! }
Thus, ( λ n n ) ≤ Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ n ≥ p > ⌊ λ n ⌋ { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } .
Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } = Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } since all prime numbers in n ! do not appear in the range of λ n ≥ p > n .
Referring to (1.5), Γ n ≥ p > ⌊ λ n ⌋ { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ≤ ∏ n ≥ p p . It has been proven [
Referring to (1.4) and (1.6), Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ≤ ( λ n ) λ n 3 + 2 .
Thus, for λ ≥ 3 and n ≥ 3 , ( λ n n ) < Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ⋅ 2 2 n − 3 ⋅ ( λ n ) λ n 3 + 2
(1.7)
Lemma 1: If a real number x ≥ 3 , then 2 ( 2 x − 1 ) x − 1 > ( x x − 1 ) x (2.1)
Proof:
Let f 1 ( x ) = 2 ( 2 x − 1 ) x − 1 ; then, f ′ 1 ( x ) = 2 ( x − 1 ) ( 2 x − 1 ) ′ − 2 ( 2 x − 1 ) ( x − 1 ) ′ ( x − 1 ) 2 = − 2 ( x − 1 ) 2 < 0 .
Thus, f 1 ( x ) is a strictly decreasing function for x > 1 .
Since f 1 ( 3 ) = 5 , and lim x → ∞ f 1 ( x ) = 4 , for x ≥ 3 , we have 5 ≥ f 1 ( x ) = 2 ( 2 x − 1 ) x − 1 ≥ 4 .
Let f 2 ( x ) = ( x x − 1 ) x , then f ′ 2 ( x ) = ( ( x x − 1 ) x ) ′ = ( e x ⋅ ln x x − 1 ) ′ = e x ⋅ ln x x − 1 ⋅ ( x ⋅ ln x x − 1 ) ′ f ′ 2 ( x ) = ( x x − 1 ) x ⋅ ( ln x x − 1 + x ⋅ ( ln x x − 1 ) ′ ) = ( x x − 1 ) x ⋅ ( ln x x − 1 + x ⋅ x − 1 x ⋅ x − 1 − x ( x − 1 ) 2 )
f ′ 2 ( x ) = ( x x − 1 ) x ⋅ ( ln x x − 1 − 1 x − 1 ) (2.1.1)
In (2.1.1), for x ≥ 3 , 1 x − 1 = 1 x + 1 x 2 + 1 x 3 + 1 x 4 + 1 x 5 + 1 x 6 + ⋯
Using the formula: ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + x 5 5 − x 6 6 + ⋯ , we have ln x x − 1 = ln 1 1 + − 1 x = − ln ( 1 + − 1 x ) = 1 x + 1 2 x 2 + 1 3 x 3 + 1 4 x 4 + 1 5 x 5 + 1 6 x 6 + ⋯
Thus, for x ≥ 3 , ln x x − 1 − 1 x − 1 < 0 .
Since ( x x − 1 ) x is a positive number for x ≥ 3 , f ′ 2 ( x ) = ( x x − 1 ) x ⋅ ( ln x x − 1 − 1 x − 1 ) < 0 .
Thus f 2 ( x ) is a strictly deceasing function for x ≥ 3 .
Since f 2 ( 3 ) = 3.375 and lim x → ∞ f 2 ( x ) = e ≈ 2.718 , for x ≥ 3 ,
3.375 ≥ f 2 ( x ) = ( x x − 1 ) x ≥ e (2.1.2)
Since for x ≥ 3 , f 1 ( x ) has a lower bound of 4 and f 2 ( x ) has an upper bound of 3.375, f 1 ( x ) = 2 ( 2 x − 1 ) x − 1 > f 2 ( x ) = ( x x − 1 ) x is proven. (2.1.3)
Lemma 2: For n ≥ 2 and λ ≥ 3 , ( λ n n ) > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 (2.2)
Proof:
When λ ≥ 3 and n = 2 , ( λ n n ) = ( 2 λ 2 ) = 2 λ ( 2 λ − 1 ) ( 2 λ − 2 ) ! 2 ( 2 λ − 2 ) ! = λ ( 2 λ − 1 )
(2.2.1)
λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 = λ 2 λ − λ + 1 2 ( λ − 1 ) 2 ( λ − 1 ) − λ + 1 = λ ( λ − 1 ) 2 ⋅ ( λ λ − 1 ) λ (2.2.2)
In (2.1) when x = λ ≥ 3 , we have 2 ( 2 λ − 1 ) λ − 1 > ( λ λ − 1 ) λ (2.2.3)
Since λ ( λ − 1 ) 2 is a positive number for λ ≥ 3 , referring to (2.2.1) and (2.2.2), when λ ( λ − 1 ) 2 multiplies both sides of (2.2.3), we have ( λ ( λ − 1 ) 2 ) ( 2 ( 2 λ − 1 ) λ − 1 ) = λ ( 2 λ − 1 ) = ( λ n n ) > ( λ ( λ − 1 ) 2 ) ( λ λ − 1 ) λ = λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 .
Thus, ( λ n n ) > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 when λ ≥ 3 and n = 2 . (2.2.4)
By induction on n, when λ ≥ 3 , if ( λ n n ) > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 is true for n, then for n + 1 ,
( λ ( n + 1 ) n + 1 ) = ( λ n + λ n + 1 ) = ( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + 1 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 ) ( n + 1 ) ⋅ ( λ n n )
( λ ( n + 1 ) n + 1 ) > ( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + 1 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 ) ( n + 1 ) ⋅ λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1
( λ ( n + 1 ) n + 1 ) > ( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 )
⋅ λ n + 1 n ⋅ 1 n + 1 ⋅ λ λ n − λ + 1 ( λ − 1 ) ( λ − 1 ) n − λ + 1
Notice λ n + 1 n > λ , and
( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 ) > ( λ λ − 1 ) λ − 1
because λ n + λ λ n + λ − n − 1 = λ λ − 1 ; λ n + λ − 1 λ n + λ − n − 2 > λ λ − 1 ; ⋯ ; λ n + 2 λ n − n + 1 > λ λ − 1 .
Thus, ( λ ( n + 1 ) n + 1 ) > λ λ − 1 ( λ − 1 ) λ − 1 ⋅ λ 1 ⋅ 1 n + 1 ⋅ λ λ n − λ + 1 ( λ − 1 ) ( λ − 1 ) n − λ + 1 .
λ λ − 1 ( λ − 1 ) ( λ − 1 ) · λ 1 · 1 ( n + 1 ) · λ λ n − λ + 1 ( λ − 1 ) ( λ − 1 ) n − λ + 1 = λ λ ( n + 1 ) − λ + 1 ( n + 1 ) ( λ − 1 ) ( λ − 1 ) ( n + 1 ) − λ + 1 .
Hence, ( λ ( n + 1 ) n + 1 ) > λ λ ( n + 1 ) − λ + 1 ( n + 1 ) ( λ − 1 ) ( λ − 1 ) ( n + 1 ) − λ + 1 (2.2.5)
From (2.2.4) and (2.2.5), we have for n ≥ 2 and λ ≥ 3 , ( λ n n ) > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1
Thus, Lemma 2 is proven.
Proposition:
For n ≥ λ − 2 ≥ 25 , there exists at least a prime number p such that
( λ − 1 ) n < p ≤ λ n . (3.1)
Proof:
Referring to (1.7), when n ≥ ( λ − 2 ) ≥ 3 , if there is a prime number p in Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } , then p ≥ n + 1 = ( n + 2 ) n + 1 > λ n . From (1.5), 0 ≤ v p ( Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ) ≤ R ( p ) ≤ 1 . Then every prime number in Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } has a power of 0 or 1. (3.2)
From (1.7), for λ ≥ 3 and n ≥ 3 , ( λ n n ) < Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ⋅ 2 2 n − 3 ⋅ ( λ n ) λ n 3 + 2 .
Applying this inequality to (2.2), when n ≥ ( λ − 2 ) ≥ 3 , λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 < ( λ n n ) < Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ⋅ 2 2 n − 3 ⋅ ( λ n ) λ n 3 + 2 . λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 < Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ⋅ 2 2 n − 3 ⋅ ( λ n ) λ n 3 + 2 . Since ( λ n ) λ n 3 + 2 > 1 and 2 2 n − 3 > 1 , Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > λ λ n − λ + 1 ( λ n ) λ n 3 + 2 ⋅ 2 2 n − 3 ⋅ n ( λ − 1 ) ( λ − 1 ) n − λ + 1 .
λ λ n − λ + 1 ( λ n ) λ n 3 + 2 ⋅ 2 2 n − 3 ⋅ n ( λ − 1 ) ( λ − 1 ) n − λ + 1 = 2 λ 2 ⋅ ( ( λ − 1 4 ) ⋅ ( λ λ − 1 ) λ ) n − 1 ( λ n ) λ n 3 + 3 .
Thus, Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 2 λ 2 ⋅ ( ( λ − 1 4 ) ⋅ ( λ λ − 1 ) λ ) ( n − 1 ) ( λ n ) λ n 3 + 3 .
Referring to (2.1.2), when λ ≥ 3 , ( λ λ − 1 ) λ ≥ e . Thus, when n ≥ ( λ − 2 ) ≥ 3 , Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 2 λ 2 ⋅ ( ( λ − 1 4 ) ⋅ ( λ λ − 1 ) λ ) n − 1 ( λ n ) λ n 3 + 3 .
2 λ 2 ⋅ ( ( λ − 1 4 ) ⋅ ( λ λ − 1 ) λ ) n − 1 ( λ n ) λ n 3 + 3 ≥ 2 λ 2 ⋅ ( ( λ − 1 4 ) ⋅ e ) n − 1 ( λ n ) λ n 3 + 3 = f 3 ( n , λ )
Thus, Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 2 λ 2 ⋅ ( ( λ − 1 4 ) ⋅ e ) ( n − 1 ) ( λ n ) λ n 3 + 3 = f 3 ( n , λ ) (3.3)
Let x ≥ 3 and y ≥ 5 both be real numbers.
When x = y – 2 , f 3 ( x , y ) = 2 ( x + 2 ) 2 ⋅ ( ( x + 1 4 ) ⋅ e ) x − 1 ( ( x + 2 ) ⋅ x ) x ⋅ ( x + 2 ) 3 + 3 > f 4 ( x ) = 2 ( x + 2 ) 2 ⋅ ( ( x + 1 4 ) ⋅ e ) x − 1 ( ( x + 2 ) ⋅ x ) x + 1 3 + 3 > 0 (3.4)
f ′ 4 ( x ) = f 4 ( x ) ⋅ ( 2 x + 2 + ln ( x + 1 4 ) + 4 3 − 2 x + 1 − 1 3 ln ( ( x + 2 ) ⋅ x ) − 10 3 x − 8 3 ( x + 2 ) ) = f 4 ( x ) ⋅ f 5 ( x )
where f 5 ( x ) = 2 x + 2 + ln ( x + 1 4 ) + 4 3 − 2 x + 1 − 1 3 ln ( ( x + 2 ) ⋅ x ) − 10 3 x − 8 3 ( x + 2 ) f 5 ( x ) = 2 x + 2 + ln ( x + 1 4 ) + 4 3 − 2 x + 1 − ln ( x ) 3 − ln ( x + 2 ) 3 − 10 3 x − 8 3 ( x + 2 ) f ′ 5 ( x ) = − 2 ( x + 2 ) 2 + 1 x + 1 + 2 ( x + 1 ) 2 − 1 3 x − 1 3 ( x + 2 ) + 10 3 x 2 + 8 3 ( x + 2 ) 2 = − 2 x 2 − 4 x − 2 ( x + 1 ) 2 ⋅ ( x + 2 ) 2 + 2 x 2 + 8 x + 8 ( x + 1 ) 2 ⋅ ( x + 2 ) 2 + 3 x 2 + 6 x 3 x ( x + 1 ) ( x + 2 ) − x 2 + 3 x + 2 3 x ( x + 1 ) ( x + 2 ) − x 2 + x 3 x ( x + 1 ) ( x + 2 ) + 10 3 x 2 + 8 3 ( x + 2 ) 2 f ′ 5 ( x ) = 4 x + 6 ( x + 1 ) 2 ⋅ ( x + 2 ) 2 + x 2 + 2 x − 2 3 x ( x + 1 ) ( x + 2 ) + 10 3 x 2 + 8 3 ( x + 2 ) 2 > 0 when x ≥ 3 .
Thus, f 5 ( x ) is a strictly increasing function for x ≥ 3 .
When x = 9 , f 5 ( x ) = 2 9 + 2 + ln ( 9 + 1 4 ) + 4 3 − 2 9 + 1 − 1 3 ln ( 9 ) − 1 3 ln ( 9 + 2 ) − 10 27 − 8 33 > 0 . Thus, for x ≥ 9 , f 5 ( x ) > 0 . Then, f ′ 4 ( x ) = f 4 ( x ) ⋅ f 5 ( x ) > 0 .
Thus, f 4 ( x ) is a strictly increasing function for x ≥ 9 .
Let x 1 = 9 and y 1 = 11 . From (3.4), when x = y – 2 , f 3 ( x , y ) > f 4 ( x ) > 0 .
Thus, when x = ( y − 2 ) ≥ 9 , then x y ≥ x 1 y 1 = 99 , f 3 ( x , y ) is an increasing function with respect to the product of x y . (3.5)
∂ f 3 ( x , y ) ∂ x = f 3 ( x , y ) ⋅ ( ln ( y − 1 4 ) + 1 − y 6 x ⋅ ln ( y x ) − y 3 x − 3 x ) = f 3 ( x , y ) ⋅ f 6 ( x , y ) (3.6)
where f 6 ( x , y ) = ln ( y − 1 4 ) + 1 − y 6 x ⋅ ln ( y x ) − y 3 x − 3 x
When x = y – 2 , then f 6 ( x , y ) = f 7 ( x ) = ln ( x + 1 4 ) + 1 − x + 2 6 x ⋅ ( ln ( x + 2 ) + ln ( x ) + 2 ) − 3 x f ′ 7 ( x ) = 1 x + 1 − x + 2 6 x ⋅ ( 1 x + 2 + 1 x ) + ln ( x + 2 ) + ln ( x ) + 2 6 x x ( x + 2 ) + 3 x 2
When x ≥ 3 , f ′ 7 ( x ) = ( 1 x + 1 − 1 3 x ( x + 2 ) ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 > 0 .
Thus, when x ≥ 3 , f 7 ( x ) is a strictly increasing function. (See Appendix A2 for more details)
When x = ( y − 2 ) ≥ 3 , since f 6 ( x , y ) = f 7 ( x ) , f 6 ( x , y ) is an increasing function respect to x y .
When x = ( y − 2 ) = 9 , f 6 ( x , y ) = ln ( 11 − 1 4 ) + 1 − 11 6 9 ⋅ ln ( 99 ) − 11 3 9 − 3 9 > 0 .
∂ f 6 ( x , y ) ∂ x = y 12 x x ⋅ ln ( y ) + y 12 x x ⋅ ln ( x ) + y 6 x x + y 6 x x + 3 x 2 > 0 when x ≥ ( y − 2 ) ≥ 3 .
Thus, when x ≥ ( y − 2 ) ≥ 9 , f 6 ( x , y ) > 0 , and it is an increasing function with respect to x and to the product of x y , then, ∂ f 3 ( x , y ) ∂ x = f 3 ( x , y ) ⋅ f 6 ( x , y ) > 0 .
Thus, when x ≥ y − 2 ≥ 9 , f 3 ( x , y ) is an increasing function with respect to x. (3.7)
Referring to (3.5) and (3.7), when x ≥ y − 2 ≥ 9 , then x y ≥ x 1 y 1 = 99 , f 3 ( x , y ) is an increasing function with respect to the product of x y . (3.8)
Let x = n and y = λ . Then when n ≥ ( λ − 2 ) ≥ 9 , f 3 ( n , λ ) is an increasing function with respect to the product of λ n and n. (3.9)
When n = ( λ − 2 ) = 25 , f 3 ( n , λ ) = 2 λ 2 ⋅ ( ( λ − 1 4 ) ⋅ e ) n − 1 ( λ n ) λ n 3 + 3 = 2 ⋅ 27 2 ⋅ ( ( 27 − 1 4 ) ⋅ e ) 25 − 1 ( 27 ⋅ 25 ) 27 ⋅ 25 3 + 3 ≈ 1.249 E + 33 9.784 E + 32 > 1 .
Since f 3 ( n , λ ) is an increasing function of the product of λ n , when n = λ − 2 ≥ 25 , f 3 ( n , λ ) > 1 .
Since f 3 ( n , λ ) is an increasing function with respect to n, when n ≥ λ − 2 ≥ 25 , f 3 ( n , λ ) > 1 .
Thus, referring to (3.3), when n ≥ λ − 2 ≥ 25 , Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > f 3 ( n , λ ) > 1 .
Let integer m ≥ n . When m ≥ n ≥ λ – 2 ≥ 25 ,
Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > f 3 ( m , λ ) > 1 . (3.10)
Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = Γ λ m ≥ p > ( λ − 1 ) m { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ ∏ i = 1 i = λ − 2 ( Γ ( λ − 1 ) m i ≥ p > λ m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m i + 1 ≥ p > ( λ − 1 ) m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } )
In ∏ i = 1 λ − 2 ( Γ ( λ − 1 ) m i ≥ p > λ m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) , for every distinct prime number p in these ranges, the numerator ( λ m ) ! has the product of p ⋅ 2 p ⋅ 3 p ⋯ i p = i ! ⋅ p i . The denominator ( ( λ − 1 ) m ) ! also has the same product of i ! ⋅ p i . Thus, they cancel each other in ( λ m ) ! ( ( λ − 1 ) m ) ! .
Referring to (1.2), ∏ i = 1 λ − 2 ( Γ ( λ − 1 ) m i ≥ p > λ m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) = 1 .
Thus, Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = Γ λ m ≥ p > ( λ − 1 ) m { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ ∏ i = 1 λ − 2 ( Γ λ m i + 1 ≥ p > ( λ − 1 ) m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } )
Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = ∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) . (3.11)
(See Appendix A3 for more details)
∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) is the product of ( λ − 1 ) sectors from i = 1 to i = ( λ − 1 ) .
Each of these sectors is the prime number factorization of the product of the consecutive integers between ( λ − 1 ) m i and λ m i .
From (3.10) and (3.11), when m ≥ n ≥ λ − 2 ≥ 25 , ∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) > 1 .
Referring to (1.1), Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ≥ 1 . Thus, when m ≥ n ≥ λ − 2 ≥ 25 , at least one of the sectors in ∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) > 1 .
Let Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } > 1 be such a sector and let m = n i where ( λ − 1 ) ≥ i ≥ 1 from (3.11). Thus, when m = n i ≥ n ≥ λ − 2 ≥ 25 ,
Γ λ n i i ≥ p > ( λ − 1 ) n i i { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } = Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } > 1 . (3.12)
( λ n i ) ! ( ( λ − 1 ) n i ) ! = ( λ n i ) ⋅ ( λ n i − 1 ) ⋯ ( λ n i − i ) ⋯ ( λ n i − 2 i ) ⋯ ( λ n i − ( n − 1 ) i ) ⋯ ( λ n i − n i + 1 ) ⋅ ( ( λ − 1 ) n i ) ! ( ( λ − 1 ) n i ) !
( λ n i ) ! ( ( λ − 1 ) n i ) ! = i ⋅ ( λ n ) ⋅ ( λ n i − 1 ) ⋯ i ⋅ ( λ n − 1 ) ⋯ i ⋅ ( λ n − 2 ) ⋯ i ⋅ ( λ n − n + 1 ) ⋯ ( λ n i − n i + 1 ) ⋅ ( ( λ − 1 ) n i ) ! ( ( λ − 1 ) n i ) !
Thus, ( λ n i ) ! ( ( λ − 1 ) n i ) ! contains all the factors of ( λ n ) , ( λ n − 1 ) , ( λ n − 2 ) , ⋯ , ( λ n − n + 1 ) in ( λ n ) ! ( ( λ − 1 ) n ) ! .
These factors make up all the consecutive integers in the range of λ n ≥ p > ( λ − 1 ) n in ( λ n ) ! ( ( λ − 1 ) n ) ! . Thus, ( λ n i ) ! ( ( λ − 1 ) n i ) ! contains ( λ n ) ! ( ( λ − 1 ) n ) ! .
Referring to the definition, all prime numbers in ( λ n i ) ! ( ( λ − 1 ) n i ) ! in the ranges of λ n i ≥ p > λ n and ( λ − 1 ) n > p do not contribute to Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } , nor does i for ( λ − 1 ) ≥ i ≥ 1 . Only the prime numbers in the prime factorization of ( λ n i ) ! ( ( λ − 1 ) n i ) ! in the range of λ n ≥ p > ( λ − 1 ) n present in Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } . Since ( λ n ) ! ( ( λ − 1 ) n ) ! is the product of all the consecutive integers in this range, Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } = Γ λ n ≥ p > ( λ − 1 ) n { ( λ n ) ! ( ( λ − 1 ) n ) ! } .
Referring to (3.12), when m = n i ≥ n ≥ λ − 2 ≥ 25 , Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } > 1 . Thus, when n ≥ λ − 2 ≥ 25 , Γ λ n ≥ p > ( λ − 1 ) n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 1 . Referring to (1.3), there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .
Thus, Proposition (3.1) is proven. It becomes a theorem: Theorem (3.1).
Legendre’s conjecture states that there is a prime number between n 2 and ( n + 1 ) 2 for every positive integer n. (4.1)
Proof:
Referring to Theorem (3.1), for integers j ≥ k − 2 ≥ 25 , there exists at least a prime number p such that j ( k − 1 ) < p ≤ j k . (4.2)
When k = j + 1 ≥ 27 , then j = k − 1 ≥ 26 ,
Applying k = j + 1 into (4.2), then j 2 < p ≤ j ( j + 1 ) < ( j + 1 ) 2 .
Let n = j ≥ 26 , then we have n 2 < p < ( n + 1 ) 2 . (4.3)
For 1 ≤ n ≤ 26 , we have a table,
Combining (4.3) and (4.4), we have proven Legendre’s conjecture.
Extension of Legendre’s conjecture
There are at least two prime numbers, p n and p m , between j 2 and ( j + 1 ) 2 for every positive integer j such that j 2 < p n ≤ j ( j + 1 ) and
j ( j + 1 ) < p m < ( j + 1 ) 2 where p n is the nth prime number, p m is the mth prime number, and m ≥ n + 1 . (4.5)
Proof:
Referring to Theorem (3.1), for integers j ≥ k − 2 ≥ 25 , there exists at least a prime number p such that j ( k − 1 ) < p ≤ j k .
When k − 1 = j ≥ 26 , then j ( k − 1 ) = j 2 < p n ≤ j k = j ( j + 1 ) . Thus, there is at least a prime number p n such that j 2 < p n ≤ j ( j + 1 ) when j = k − 1 ≥ 26 .
When j = k − 2 ≥ 26 , then k = j + 2 .
j ( k − 1 ) = j ( j + 1 ) < p m ≤ j k = j ( j + 2 ) < ( j + 1 ) 2 . Thus, there is at least another prime number p m such that j ( j + 1 ) < p m < ( j + 1 ) 2 when j = k − 2 ≥ 26 .
Thus, when j ≥ 26 , there are at least two prime numbers p n and p m between j 2 and ( j + 1 ) 2 such that j 2 < p n ≤ j ( j + 1 ) < p m < ( j + 1 ) 2 where
m ≥ n + 1 for p m > p n . (4.6)
For 1 ≤ j ≤ 26 , we have a table,
Combining (4.6) and (4.7), we have proven (4.5). It becomes a theorem: Theorem (4.5).
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| n2 | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 |
| p | 3 | 5 | 11 | 19 | 29 | 41 | 53 | 67 | 83 | 103 | 127 | 149 | 173 |
| (n + 1)2 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 | 196 |
| n | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 |
| n2 | 196 | 225 | 256 | 289 | 324 | 361 | 400 | 441 | 484 | 529 | 576 | 625 | 676 |
| p | 199 | 229 | 263 | 307 | 331 | 373 | 409 | 449 | 491 | 541 | 587 | 641 | 683 |
| (n + 1)2 | 225 | 256 | 289 | 324 | 361 | 400 | 441 | 484 | 529 | 576 | 625 | 676 | 729 |
| j | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| j2 | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 |
| pn | 2 | 5 | 11 | 19 | 29 | 41 | 53 | 67 | 83 | 103 | 127 | 149 | 173 |
| j(j + 1) | 2 | 6 | 12 | 20 | 30 | 42 | 56 | 72 | 90 | 110 | 132 | 156 | 182 |
| pm | 3 | 7 | 13 | 23 | 31 | 43 | 59 | 73 | 97 | 113 | 137 | 163 | 191 |
| (j + 1)2 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 | 196 |
| j | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 |
| j2 | 196 | 225 | 256 | 289 | 324 | 361 | 400 | 441 | 484 | 529 | 576 | 625 | 676 |
| pn | 199 | 229 | 263 | 293 | 331 | 373 | 409 | 449 | 491 | 541 | 587 | 641 | 683 |
| j(j + 1) | 210 | 240 | 272 | 306 | 342 | 380 | 420 | 462 | 506 | 552 | 600 | 650 | 702 |
| pm | 211 | 251 | 277 | 311 | 349 | 389 | 431 | 467 | 521 | 557 | 613 | 659 | 709 |
| (j + 1)2 | 225 | 256 | 289 | 324 | 361 | 400 | 441 | 484 | 529 | 576 | 625 | 676 | 729 |
Oppermann’s conjecture was proposed in March 1877 by Ludvig Oppermann (1817-1883). It states that for every integer x > 1 , there is at least one prime number between x ( x − 1 ) and x 2 , and at least another prime number between x 2 and x ( x + 1 ) . [
Proof:
Theorem (4.5) states that there are at least two prime numbers, p n and p m , between j 2 and ( j + 1 ) 2 for every positive integer j such that j 2 < p n ≤ j ( j + 1 ) < p m < ( j + 1 ) 2 where m ≥ n + 1 for p m > p n .
j ( j + 1 ) is a composite number except j = 1 . Since j 2 < p n ≤ j ( j + 1 ) is valid for every positive integer j, when we replace j with j + 1 , we have ( j + 1 ) 2 < p v < ( j + 1 ) ( j + 2 ) .
Thus, we have j ( j + 1 ) < p m < ( j + 1 ) 2 < p v < ( j + 1 ) ( j + 2 ) . (5.2)
When x > 1 , then ( x − 1 ) ≥ 1 . Substituting j with ( x − 1 ) in (5.2), we have
x ( x − 1 ) < p m < x 2 < p v < x ( x + 1 ) (5.3)
Thus, we have proven Oppermann’s conjecture.
Brocard’s conjecture is based on Henri Brocard (1845-1922). It states that there are at least 4 prime numbers between ( p n ) 2 and ( p n + 1 ) 2 , where p n is the nth prime number, for every n > 1 . [
Proof:
Theorem (4.5) states that there are at least two prime numbers, p n and p m , between j 2 and ( j + 1 ) 2 such that j 2 < p n ≤ j ( j + 1 ) and j ( j + 1 ) < p m < ( j + 1 ) 2 for every positive integer j, where m ≥ n + 1 for p m > p n . When j > 1 , j ( j + 1 ) is a composite number. Then Theorem (4.5) can be written as j 2 < p n < j ( j + 1 ) and j ( j + 1 ) < p m < ( j + 1 ) 2 .
In the series of prime numbers: p 1 = 2 , p 2 = 3 , p 3 = 5 , p 4 = 7 , p 5 = 11 , ... all prime numbers except p 1 are odd numbers. Their gaps are two or more. Thus, when n > 1 , ( p n + 1 − p n ) ≥ 2 . Thus, we have p n < ( p n + 1 ) < ( p n + 2 ) ≤ p n + 1 when n > 1 . (5.5)
Applying Theorem (4.5) to (5.5), when n > 1 , we have at least two prime numbers p m 1 , and p m 2 in between ( p n ) 2 and ( p n + 1 ) 2 such that ( p n ) 2 < p m 1 < p n ( p n + 1 ) < p m 2 < ( p n + 1 ) 2 , and at least two more prime numbers p m 3 , p m 4 in between ( p n + 1 ) 2 and ( p n + 2 ) 2 such that ( p n + 1 ) 2 < p m 3 < ( p n + 1 ) ( p n + 2 ) < p m 4 < ( p n + 2 ) 2 ≤ ( p n + 1 ) 2 .
Thus, there are at least 4 prime numbers between ( p n ) 2 and ( p n + 1 ) 2 for n > 1 such that
( p n ) 2 < p m 1 < p n ( p n + 1 ) < p m 2 < ( p n + 1 ) 2 < p m 3 < ( p n + 1 ) ( p n + 2 ) < p m 4 < ( p n + 1 ) 2 (5.6)
Thus, Brocard’s conjecture is proven.
Andrica’s conjecture is named after Dorin Andrica [
g n = p n + 1 − p n denotes the nth prime gap, then Andrica’s conjecture can also be rewritten as g n < 2 p n + 1 . (5.7)
Proof:
From Theorem (4.5), for every positive integer j, there are at least two prime numbers p n and p m between j 2 and ( j + 1 ) 2 such that j 2 < p n ≤ j ( j + 1 ) < p m < ( j + 1 ) 2 where m ≥ n + 1 for p m > p n . Since m ≥ n + 1 , we have p m ≥ p n + 1 .
Thus, we have j 2 < p n (5.8)
and p n + 1 ≤ p m < ( j + 1 ) 2 . (5.9)
Since j, p n , p n + 1 and ( j + 1 ) are positive integers,
j < p n (5.10)
and p n + 1 < j + 1 (5.11)
Applying (5.10) to (5.11), we have p n + 1 < p n + 1 . (5.12)
Thus, p n + 1 − p n < 1 holds for all n since in Theorem (4.5), j holds for all positive integers.
Using the prime gap to prove the conjecture, from (5.8) and (5.9), we have g n = p n + 1 − p n < ( j + 1 ) 2 − j 2 = 2 j + 1 . From (5.10), j < p n .
Thus, g n = p n + 1 − p n < 2 p n + 1 . (5.13)
Thus, Andrica’s conjecture is proven.
The author declares no conflicts of interest regarding the publication of this paper.
Yu, W.K. (2023) The Proofs of Legendre’s Conjecture and Three Related Conjectures. Journal of Applied Mathematics and Physics, 11, 1319-1336. https://doi.org/10.4236/jamp.2023.115085
At the 1912 International Congress of Mathematicians, Edmund Landau listed four basic problems about primes. These problems were characterized in his speech as “unattackable at the present state of mathematics” and are now known as Landau’s problems. They are as follows:
1) Goldbach’s conjecture: Can every even integer greater than 2 be written as the sum of two primes?
2) Twin prime conjecture: Are there infinitely many primes p such that p + 2 is prime?
3) Legendre’s conjecture: Does there always exist at least one prime between consecutive perfect squares?
4) Are there infinitely many primes p such that p – 1 is a perfect square? In other words: Are there infinitely many primes of the form m2 + 1?
As of 2023, all four problems are unresolved.
The above content is quoted from Wikipedia. https://en.wikipedia.org/wiki/Landau%27s_problems
The author of this paper has browsed the articles about the proof of Legendre’s conjecture posted on and published in different media in recent years, and made some brief comments, as below.
1) Kouji Takaki, Proof of Legendre’s Conjecture, (Mar. 2023) https://vixra.org/abs/2110.0102
Comment: Using the empirical method to prove Legendre’s conjecture is considered insufficient evidence.
2) Zhi Li and Hua Li, Proof of Legendre Conjecture, (Sep. 2022) https://vixra.org/pdf/2209.0070v1.pdf
Comment: The method of probability and statistics introduces the uncertainty in a certain range. It is not the right choice to prove Legendre’s conjecture.
3) Kouider Mohammed Ridha, On Legendre’s Conjecture, (Sep. 2022) https://vixra.org/pdf/2209.0051.pdf
Comment: The prime number formular p ( n ) = n 2 + n − 1 is not correct.
4) Kenneth Watanabe, Definitive Proof of Legendre’s Conjecture, (2019) https://vixra.org/pdf/1901.0436v1.pdf
Comment: The asymptotic method is not the right choice to prove Legendre’s conjecture.
5) Ahmed Telfah, A Proof of Legendre’s Conjecture and Andrica’s Conjecture, (Dec. 2018) https://www.researchgate.net/publication/329844915_A_proof_of_Legendre’s_conjecture_and_Andrica’s_conjecture
Comment: Using the distribution function to deal with discrete primes creates bound problems. The empirical method is not sufficient to solve these problems.
6) Samuel Bonaya Buya, Proof of Legendre’s Conjecture, Research & Reviews: Journal of Applied Science and Innovations, RRJASI, Volum2, Issue 2, January-March, 2018
Comment: There is a contradiction between Equation (2) and Equation (9), which is a fatal error.
7) Samuel Bonaya Buya, A Simple Proof of Legendre’s Conjecture, (2018) https://www.academia.edu/35702628/A_simple_proof_of_Legendres_conjecture
Comment: Using the prime number function to estimate the prime number between n 2 and ( n + 1 ) 2 is not the right way to prove Legendre’s conjecture.
A2Derivation details for f ′ 7 ( x ) > 0
When x = y − 2 ≥ 3 , then
f 6 ( x , y ) = f 7 ( x ) = ln ( x + 1 4 ) + 1 − x + 2 6 x ⋅ ( ln ( x + 2 ) + ln ( x ) + 2 ) − 3 x
d ( x + 2 6 x ) d x = 1 6 x ⋅ x 2 x + 2 − 1 6 x ⋅ x + 2 2 x = x − x − 2 12 x x ( x + 2 ) = − 1 6 x x ( x + 2 )
f ′ 7 ( x ) = 1 x + 1 − x + 2 6 x ⋅ ( 1 x + 2 + 1 x ) + ln ( x + 2 ) + ln ( x ) + 2 6 x x ( x + 2 ) + 3 x 2 = 1 x + 1 − x + 2 6 x ⋅ x + x + 2 x ( x + 2 ) + ln ( x + 2 ) + ln ( x ) + 2 6 x x ( x + 2 ) + 3 x 2 = 1 x + 1 − 1 3 x ⋅ x + 1 x x + 2 + 2 6 x x ( x + 2 ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 = 1 x + 1 − x + 1 3 x x ( x + 2 ) + 1 3 x x ( x + 2 ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 = 1 x + 1 − x 3 x x ( x + 2 ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 = ( 1 x + 1 − 1 3 x ( x + 2 ) ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2
When x ≥ 1 , then 3 x ( x + 2 ) + ( x + 1 ) > 0
( 3 x ( x + 2 ) + ( x + 1 ) ) ⋅ ( 3 x ( x + 2 ) − ( x + 1 ) ) = ( 3 x ( x + 2 ) ) 2 − ( x + 1 ) 2 = 8 x 2 + 16 x − 1 > 0
Thus, ( 3 x ( x + 2 ) + ( x + 1 ) ) ⋅ ( 3 x ( x + 2 ) − ( x + 1 ) ) > 0
3 x ( x + 2 ) − ( x + 1 ) > 0
3 x ( x + 2 ) > ( x + 1 ) then 1 x + 1 > 1 3 x ( x + 2 )
When x ≥ 1 , ( 1 x + 1 − 1 3 x ( x + 2 ) ) > 0 , and ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 > 0 .
Then f ′ 7 ( x ) = ( 1 x + 1 − 1 3 x ( x + 2 ) ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 > 0 .
Thus, when x ≥ 3 , f 7 ( x ) is a strictly increasing function.
A3Derivation details for (3.11)
Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = Γ λ m ≥ p > ( λ − 1 ) m { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ ( λ − 1 ) m ≥ p > λ m 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m 2 ≥ p > ( λ − 1 ) m 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ ( λ − 1 ) m 2 ≥ p > λ m 3 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m 3 ≥ p > ( λ − 1 ) m 3 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ ( λ − 1 ) m 3 ≥ p > λ m 4 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋯ Γ ( λ − 1 ) m λ − 3 ≥ p > λ m λ − 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m λ − 2 ≥ p > ( λ − 1 ) m λ − 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ ( λ − 1 ) m λ − 2 ≥ p > λ m λ − 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m λ − 1 ≥ p > ( λ − 1 ) m λ − 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! }
In Γ ( λ − 1 ) m ≥ p > λ m 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } , for every distinct prime number p in this range,
the numerator ( λ m ) ! has the prime number p. The denominator ( ( λ − 1 ) m ) !
also has the same p. Thus, they cancel each other in ( λ m ) ! ( ( λ − 1 ) m ) ! . Referring to (1.2), Γ ( λ − 1 ) m ≥ p > λ m 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } = 1 .
In Γ ( λ − 1 ) m 2 ≥ p > λ m 3 { ( λ m ) ! ( ( λ − 1 ) m ) ! } , for every distinct prime number p in this range, the numerator ( λ m ) ! has the product of p ⋅ 2 p . The denominator ( ( λ − 1 ) m ) ! also has the same product of p ⋅ 2 p . Thus, they cancel each other in ( λ m ) ! ( ( λ − 1 ) m ) ! .
Referring to (1.2), Γ ( λ − 1 ) m 2 ≥ p > λ m 3 { ( λ m ) ! ( ( λ − 1 ) m ) ! } = 1 .
In Γ ( λ − 1 ) m 3 ≥ p > λ m 4 { ( λ m ) ! ( ( λ − 1 ) m ) ! } , for every distinct prime number p in this range,
the numerator ( λ m ) ! has the product of p ⋅ 2 p ⋅ 3 p . The denominator
( ( λ − 1 ) m ) ! also has the same product of p ⋅ 2 p ⋅ 3 p . Thus, they cancel each other in ( λ m ) ! ( ( λ − 1 ) m ) ! .
Referring to (1.2), Γ ( λ − 1 ) m 3 ≥ p > λ m 4 { ( λ m ) ! ( ( λ − 1 ) m ) ! } = 1 .
⋮
In Γ ( λ − 1 ) m λ − 3 ≥ p > λ m λ − 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } , for every distinct prime number p in this range,
the numerator ( λ m ) ! has the product of p ⋅ 2 p ⋅ 3 p ⋯ ( λ − 3 ) p . The denominator ( ( λ − 1 ) m ) ! also has the same product of p ⋅ 2 p ⋅ 3 p ⋯ ( λ − 3 ) p . Thus,
they cancel each other in ( λ m ) ! ( ( λ − 1 ) m ) ! .
Referring to (1.2), Γ ( λ − 1 ) m λ − 3 ≥ p > λ m λ − 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } = 1 .
In Γ ( λ − 1 ) m λ − 2 ≥ p > λ m λ − 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } , for every distinct prime number p in this range,
the numerator ( λ m ) ! has the product of p ⋅ 2 p ⋅ 3 p ⋯ ( λ − 2 ) p . The denominator ( ( λ − 1 ) m ) ! also has the same product of p ⋅ 2 p ⋅ 3 p ⋯ ( λ − 2 ) p . Thus,
they cancel each other in ( λ m ) ! ( ( λ − 1 ) m ) ! .
Referring to (1.2), Γ ( λ − 1 ) m λ − 2 ≥ p > λ m λ − 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } = 1 .
Thus,
Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = Γ λ m ≥ p > ( λ − 1 ) m { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m 2 ≥ p > ( λ − 1 ) m 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m 3 ≥ p > ( λ − 1 ) m 3 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋯ Γ λ m λ − 2 ≥ p > ( λ − 1 ) m λ − 2 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m λ − 1 ≥ p > ( λ − 1 ) m λ − 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! }
Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = ∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) . (3.11)