Sometimes it is difficult to get feedback about a calculation or the steps of a proof. This is one of those cases. This is a mirror of my
question at MSE.
Let
p_n denote the
n^{th} prime number.
Ingham showed that:
p_{n+1} - p_n \lt K p_n^{\frac{5}{8}}
where
K is a fixed positive integer, is an
upper bound for the prime gaps.
(A.E.Ingham, On the difference between consecutive primes, Quart. J. Math. Oxford Ser. vol. 8 (1937) pp. 255-266)
I am trying to know if the following manipulations would be valid to reduce the value
\frac{5}{8} up to
\frac{1}{8}. The steps
1,2,3 have been verified in, but still I am trying to get some help to review the rest of them.
1. Both sides to the eighth:
(p_{n+1} - p_n)^8 \lt K^8 p_n^{5}
2. Def.
K_2 = K^8 a new fixed constant
K_2 \gt K:
(p_{n+1} - p_n)^8 \lt K_2 p_n^{5}
3. By
Fermat's Little Theorem, we know that
p_n^{5} can be replaced as follows}:
p_n^{5}= 5K^{'} + p_n
For a given positive
unbounded constant
K^{'}. Thus:
(p_{n+1} - p_n)^8 \lt K_2 (5K^{'} + p_n)
But (credits for the following explanation to a
MSE user here) the map
K' : \mathbb{P} \to \mathbb{N} : p \mapsto \frac{p^5 - p}{5}
is not constant and unbounded from above.
\color{red}{From\ this\ point\ ahead\ the\ manipulations\ need\ to\ be\ verified}.
The right side of the inequality can be replaced as follows:
(p_{n+1} - p_n)^8 \lt K_2 (5K^{'} + p_n) = (K_2 \cdot 5K^{'}) + K_2 p_n
Our partial refinement will be true
only if the constants are fixed, so we need to add a condition to handle the unbounded constant
(K_2 \cdot 5K^{'}). So let us assume the following condition:
Condition 1:
p_n^{5}-p_n = 5K^{'} \lt K_2 = K^8
So from now on, the partial refinement will be valid only those primes
p_n whose value
p_n^{5}-p_n is bounded by the eighth power of Ingham's
K constant.
4. Def.
K_3 = K_2^2 an even bigger new fixed constant
K_3 \gt K_2:
(p_{n+1} - p_n)^8 \lt K_3 + K_2 p_n
5. As
K_3 \gt K_2, it is also true that:
(p_{n+1} - p_n)^8 \lt K_3 + K_2 p_n \lt K_3 + K_3 p_n = K_3 (1+p_n) \lt K_3 (2p_n) = (2K_3)p_n
6. Def.
K_4 = 2K_3 as an even bigger new fixed constant
K_4 \gt K_3:
(p_{n+1} - p_n)^8 \lt K_4p_n
7. Now we make again the
8^{th} root in both sides:
p_{n+1} - p_n \lt K_4^{\frac{1}{8}}p_n^{\frac{1}{8}}
8. Finally, def.
K_5 = K_4^{\frac{1}{8}} as a new fixed constant. In this case
K_5 \lt K_4 but still fixed and positive:
p_{n+1} - p_n \lt K_5p_n^{\frac{1}{8}}
Only valid under Condition 1:
p_n^{5}-p_n \lt K^8
being
K the original Ingham's
K constant.
Thus, if the manipulations and redefinition of constants are correct:
1. If
p_n^{5}-p_n \lt K^8:
p_{n+1} - p_n \lt K_5 p_n^{\frac{1}{8}} = 2^{\frac{1}{8}}K^2 p_n^{\frac{1}{8}}
for a fixed constant
K_5 = K_4^{\frac{1}{8}}=(2K_3)^{\frac{1}{8}}=(2K_2^2)^{\frac{1}{8}}=(2(K^8)^2)^{\frac{1}{8}}=2^{\frac{1}{8}}K^2, and
2. In the rest of unbounded cases remains as the original Ingham's expression:
p_{n+1} - p_n \lt K p_n^{\frac{5}{8}}
I have asked at MSE if this is correct or not, and if it is useful or not. If it is correct, is just valid for a finite range of prime numbers
p_n=2,3,5... as long as
p_n^{5}-p_n \lt K^8. If I receive an answer I will write here the results too.
