An Approach to solve Andrica's Conjecture

Question

Hi I used the following steps to solve the Andrica's conjecture.

Problem Statement : $\sqrt{p_{n+1}}$$-$$\sqrt{p_{n}}$ < 1 Where $p_n$ is the $n^{th}$ prime number

Steps: We can easily prove $(log($$p_n))^2$ < $p_n$ for all $p_n$>1

multiplying with $p_n$ on both sides

$p_n$$(log($$p_n))^2$$ < $$p_n^2$

$p_n$ < $($$p_n$/$log($$p_n$$)$$)$$^2$

From the Prime Number theorem,

($p_n$/$log($$p_n$$)$$)$ < $\pi$($p_n$)

so $p_n$ < ($\pi$($p_n$))$^2$

$p_n$ < $n^2$

$\sqrt{p_{n}}$ < n

which implies the statement $\sqrt{p_{n+1}}$$-$$\sqrt{p_{n}}$ < 1

Please anyone give some insights on this approach.

Answer 1

I'm not bothering with checking the initial calculations, but $\sqrt{p_n} < n$ does in no way imply $\sqrt{p_{n+1}}-\sqrt{p_{n}} < 1$. There is no statement about a lower bound on $\sqrt{p_n}$, so maybe $\sqrt{p_{1000}}=998.56\ldots$ and $\sqrt{p_{1001}}=1000.23\ldots$. Both values hold the bound you give, but their difference is still bigger than $1$.

Answer 2

Note: if we have $\sqrt p_n \leq n \Rightarrow -n \leq-\sqrt p_n \leq 0 $ then $ |\sqrt p_{n+1} - \sqrt p_n | \leq n+1$ ,the follwing expression is false : $\sqrt p_{n+1} - \sqrt p_n \leq n+1-n \leq 1$

Answer 3

Indeed, Andrica's conjecture (and something quite a bit stronger) would be true (for large enough $n$) if for all $n$ there is a prime in the interval $n,n+1,\ldots, n+j$ for some $j \in \theta(\log n)$. In fact, Andrica's conjecture would be true (for large enough $n$) if for all $n$ there is a prime in the interval $n,n+1,\ldots, n+j$ for some $j \in o(\log n \sqrt{n})$ i.e., $j$ doesn't have to so small to be logarithmic in size relative to $n$ before you are guaranteed to come across a prime, instead $j$ is allowed to be much larger--it can be almost as large as $\sqrt{n} \log n$.

The Prime Number Theorem does not quite give either of the above though. It does not give a precise enough bound on the number of primes less than $n$ to imply either of the above. The PNT says that there are $\frac{n(1\pm o(1))}{\log n}$ primes less than or equal to $n$ for all integers $n$. This still allows that there could be many integers $n$ such that there are no primes in $n,n+1,\ldots, n+j$ for some $j \in \omega(\log n \sqrt{n})$, and these intervals are balanced out by integers $n$ such that there are "too many" primes in $n,n+1,\ldots, n+j$ for some $j \in w(\sqrt{n})$.