Is this probabilistic proof for Brocard's Conjecture flawed?

Brocard’s conjecture is the assertion that there are at least four primes between consecutive prime squares, for primes greater than 2. In notation,

$pi(p_{n+1}^{2})-pi(p_{n}^{2})ge4$
for $ngt1$
(so $p_ngt2$).

I’ve been working on Legendre’s conjecture for a while, so I figured taking a stab at a weaker form might give me insight to a more ‘fertile’ method of investigation. Anyway, here’s what I have so far, and it feels… incomplete. If you could give it a run down, it would be much appreciated! 🙂

(Also, I’m sorry my language is a bit complicated – I need to re-study set theory :^)

All $x$ on the open interval $(1,p_{n+1}^{2})$ of the form

$x=(p_{n}#)cdot{k}+q$,

where

• $k,qin{Bbb{Z_{0}^{+}}}$
• $qlt{p_{n}#}$
• and $gcd(q,p_{n}#)=1$
• (and $#$ denotes the primorial function)

are prime (Eratosthenes).

As an example, all numbers that are coprime to $30$, which is $2cdot{3}cdot{5}$, on the open interval $(1,7^{2})$ are prime: $7,11,13,17,19,23,29,31,37,41,43,47$. This follows from the Sieve of Eratosthenes, and the fact that a proper divisor $d$ of a number $N$ must be less than or equal to the square root of that number $N$.

The only primes on the open interval $(1,p_{n+1}^{2})$ that are not of the form $x=(p_{n}#)cdot{k}+q$ are those primes

$p_{C}, Cle{n}$,

because when $k=0,x=q$, so $gcd(x,p_{n}#)ne{1}$ if $x=p_{C}$,

and when $kgt{0},x=(p_{n}#)cdot{k}+qgt{p_n}#gt{p_{C}}$, so $xne P_{C}$,

and there are $n$ of these primes.

There are $phi{(p_{n}#)}$ $q$‘s coprime to $p_{n}#$, where $phi{(x)}$ is Euler’s totient function, by definition (Euler).
A well known way to calculate the totient of a primorial is by the product

$phi(p_{n}#)=p_{n}#cdot{prodlimits_{primespace{p_i}}^{p_{n}}}{left(1-frac{1}{p_i}right)}$

Let $bar{phi}(p_{n}#)$ denote the proportion of q’s over $(1,p_{n}#)$, that is:

$bar{phi}(p_{n}#)=frac{p_{n}#cdot{prodlimits_{primespace{p_i}}^{p_{n}}}{left(1-frac{1}{p_i}right)}}{p_{n}#}=prodlimits_{primespace{p_i}}^{p_{n}}{left(1-frac{1}{p_i}right)}=prodlimits_{primespace{p_i}}^{p_{n}}{left(frac{p_i-1}{p_i}right)}$

Now, using this proportion $bar{phi}$, we can estimate the number of numbers on $(1,p_{n+1}^{2})$ that are coprime to $p_{n}#$ and thus prime, like so:

$left(bar{phi}(p_{n}#)right)cdotleft(p_{n+1}^{2}-2right)$.

This is derived from the proportion $bar{phi}$ multiplied by the amount of numbers on the interval, $left(p_{n+1}^{2}-2right)$.

And adding back the $n$ first primes $p_{C}$, we have:

$pi(p_{n+1}^2)approx{n}+left(bar{phi}(p_{n}#)right)cdotleft(p_{n+1}^{2}-2right)$. You may see where I’m going with this.

Let’s now consider the proportion $bar{phi}(p_{n}#)$:

Reorganizing, $bar{phi}(p_{n}#)=prodlimits_{primespace{p_i}}^{p_n}left(frac{p_i-1}{p_i}right)=frac{1}{p_n}cdotprodlimits_{primespace{p_i}gt2}^{p_n}left(frac{p_i-1}{p_{i-1}}right)$.

As $p_{n+1}ge{p_{n}+2}$, so $prodlimits_{primespace{p_i}gt2}^{p_n}left(frac{p_i-1}{p_{i-1}}right)gt{1}$ for $n>1$, and $bar{phi}(p_n#)gt{frac{1}{p_{n}}}gt{frac{1}{p_{n+1}}}$

For $n=4$, we have $p_{4}=7$, so

$bar{phi}(7#)=prodlimits_{prime space {p_i}}^{7}frac{p_i-1}{p_i}=frac{1}{2}cdotfrac{2}{3}cdotfrac{4}{5}cdotfrac{6}{7}=1cdotfrac{2}{2}cdotfrac{4}{3}cdotfrac{6}{5}cdotfrac{1}{7}=frac{8}{5}cdotfrac{1}{7}$

More specifically, $bar{phi}(p_{n})gefrac{8}{5}cdotfrac{1}{p_{n}}gtfrac{8}{5}cdotfrac{1}{p_{n+1}}$ when $nge4$.

Rewriting as a recurrence relationship, $bar{phi}(p_{n}#)=bar{phi}(p_{n-1}#)cdotfrac{p_{n}-1}{p_{n}}$, so $bar{phi}(p_{n}#)gtfrac{8}{5}cdotfrac{p_{n}-1}{p_{n}^2}gtfrac{8}{5}cdotfrac{p_{n+1}-1}{p_{n+1}^2}$ when $ngt4$.

Now, finally, let

$A:=pi(p_{n+1}^2)approx{n}+bar{phi}(p_{n}#)cdot{left(p_{n+1}^2-2right)}$, and

$B:=pi(p_{n}^2)approx{n-1}+bar{phi}(p_{n-1}#)cdotleft(p_{n}^2-2right)$.

Restating $A$ using the recurrence relation,

$Aapprox{n+left(frac{p_n-1}{p_n}right)cdotbar{phi}(p_{n-1}#)cdotleft(p_{n+1}^2-2right)}$.

Taking the difference, $A-Bapprox{1+bar{phi}(p_{n-1}#)cdotleft[left(frac{p_{n}-1}{p_{n}}right)cdotleft(p_{n+1}^2-2right)-left(p_{n}^2-2right)right]}$, and simplifying using $p_{n+1}ge p_{n}+2$,

$A-Bge{1+bar{phi}(p_{n-1}#)cdotleft[left(frac{p_{n}-1}{p_{n}}right)cdotleft((p_{n}+2)^2-2right)-left(p_{n}^2-2right)right]}$

Skipping some working out…(I might add a comment with this particular working out)

$A-Bge{1+bar{phi}(p_{n-1}#)cdotleft[frac{3p_{n}^2-2}{p_{n}}right]}$

And lastly, using the $bar{phi}$ inequality,

$A-Bgt{1+frac{8}{5}cdotleft(frac{p_n-1}{p_n^2}right)cdotleft[frac{3p_{n}^2-2}{p_{n}}right]}=1+frac{8}{5}cdotleft(3-frac{3}{p_{n}}-frac{2}{p_{n}^2}+frac{2}{p_{n}^3}right)gt4$ when $p_ngt3$, which agrees with the range of $nge4, p_nge{7}$ from the $bar{phi}$ inequality. The existence of at least four primes between $3^2$ and $5^2$, and at least four more between $5^2$ and $7^2$ can be verified manually $left[left(11,13,17,19,23right),left(29,31,37,41,43,47right)right]$.

Q.E.D.?