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.
