With the previous post, I stopped at the argument that f_{4}(x)=\pi (x) - \sqrt{x} seems to be ascending for prime arguments. Apparently, this is a necessary and sufficient condition, i.e.:
Lemma 1. Andrica's conjecture is true iff function f_{4}(x)=\pi (x) - \sqrt{x} is strictly ascending (x < y \Rightarrow f(x) < f(y)) for prime arguments.
The proof is given in the following discussion with the math.stackexchange.com community.
It is worth mentioning that f_{4}(x) is descending in between prime numbers. For example for \forall x\in \left ( p_{n}, p_{n+1} \right ), f_{4}(x)=n - \sqrt{x} and {f_{4}}'(x)=-\frac{1}{2\cdot \sqrt{x}} < 0.
In fact the purpose of the discussion was to prove that \sqrt{p_{n}} < n which turns to be true for \forall n \geq 2, using Rosser's theorem.
Interesting result, out of this inequality, is that \sqrt{p_{n}} - \sqrt{p_{1}} < n-1 because \sqrt{p_{1}}=\sqrt{2} > 1. However, \sum_{k=2}^{n}\left ( \sqrt{p_{k}} - \sqrt{p_{k-1}} \right )=\sqrt{p_{n}} - \sqrt{p_{1}} < n-1 and as a result \frac{1}{n-1}\cdot \sum_{k=2}^{n}\left ( \sqrt{p_{k}} - \sqrt{p_{k-1}} \right ) < 1 or, in other words, on average Andrica's conjecture seems to be true.
But, leaving the statistical argument aside, how do we prove that f_{4}(x) is ascending for prime arguments?
I don't have the answer yet, but here is where I stuck ...
Let's have a look at the following function f_{5}(x)=1 - \sqrt{p_{n+1} + x} + \sqrt{p_{n} + x} This function is ascending \forall x \geq -p_{n} {f_{5}(x)}'= -\frac{1}{2\cdot \sqrt{p_{n+1} + x}} + \frac{1}{2\cdot \sqrt{p_{n} + x}} > 0 because \sqrt{p_{n+1} + x} > \sqrt{p_{n} + x}, for \forall x \geq -p_{n}.
This function also has a zero, 1=\sqrt{p_{n+1} + x} - \sqrt{p_{n} + x} \Leftrightarrow 1 = p_{n+1}+p_{n} +2\cdot x -2\cdot \sqrt{(p_{n+1} + x)\cdot (p_{n} + x)} \Leftrightarrow 2\cdot \sqrt{(p_{n+1} + x)\cdot (p_{n} + x)}=p_{n+1}+p_{n} +2\cdot x - 1 \Leftrightarrow 4\cdot (p_{n+1} + x)\cdot (p_{n} + x)=p_{n+1}^{2}+p_{n}^{2}+4\cdot x^{2}+1+2\cdot p_{n+1}\cdot p_{n}+ +4\cdot p_{n+1}\cdot x-2\cdot p_{n+1}+4\cdot p_{n}\cdot x-2\cdot p_{n}-4\cdot x \Leftrightarrow jumping ahead 2\cdot p_{n+1}\cdot p_{n}=p_{n+1}^{2}+p_{n}^{2}+1-2\cdot p_{n+1}-2\cdot p_{n}-4\cdot x \Leftrightarrow 4\cdot x=(p_{n+1}-p_{n})^{2}-2\cdot (p_{n+1}+p_{n})+1=(p_{n+1}-p_{n}-1)^{2}-4\cdot p_{n} as a result x_{0}=\left( \frac{p_{n+1}-p_{n}-1}{2} \right)^{2}-p_{n} Obviously, x_{0} > -p_{n}.
And now:
Lemma 2. f_{4}(p_{n+1}) > f_{4}(p_{n}) iff x_{0} < 0, where x_{0} is zero of f_{5}(x).
First of all x_{0} < 0 \Leftrightarrow p_{n+1}-p_{n} < 2\cdot \sqrt{p_{n}}+1
As a result, if f_{4}(p_{n+1}) > f_{4}(p_{n}) this is equivalent (from Lemma 1) with \sqrt{p_{n+1}}-\sqrt{p_{n}} < 1 (Andrica's inequality for p_{n}, p_{n+1}). But then p_{n+1} - p_{n}=(\sqrt{p_{n+1}}-\sqrt{p_{n}})\cdot (\sqrt{p_{n+1}}+\sqrt{p_{n}}) < \sqrt{p_{n+1}}+\sqrt{p_{n}} and applying Andrica's inequality again p_{n+1} - p_{n} < \sqrt{p_{n}}+1+\sqrt{p_{n}}=2\cdot \sqrt{p_{n}}+1 so x_{0} < 0.
Now, if x_{0} < 0, where we know that x_{0} > -p_{n}, and the fact that f_{5}(x) is ascending then we have x_{0} < 0 \Rightarrow 0=f_{5}(x_{0}) < f_{5}(0)=1 - \sqrt{p_{n+1}} + \sqrt{p_{n}} or \sqrt{p_{n+1}}-\sqrt{p_{n}} < 1 and then (Lemma 1) f_{4}(p_{n+1}) > f_{4}(p_{n}).
Do we know if p_{n+1}-p_{n} < 2\cdot \sqrt{p_{n}}+1 for \forall n?
Not yet, but we are close. One result, proved by Martin Huxley, states that p_{n+1}-p_{n} < p_{n}^{\theta } for \theta > \frac{7}{12} and sufficiently large n. In our case, \theta =0.5
Another result by Baker, Harman and Pintz states that \theta may be taken to be 0.525.
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