# Math Genius: For what percentage of numbers does this proof of Goldbach’s conjecture hold?

## Question

For what percentage of numbers does the below inequality hold?

$$pi(2m) > frac{phi(2 m) -1}{2}$$

where $$m$$ is not a prime or $$1$$, $$pi(m)$$ is the number of primes less than $$m$$ and $$phi$$ is the Euler totient function.

## Background + Proof

I was trying to prove Goldbach’s Conjecture are realised I could do it for certain numbers.

Consider the following where $$m$$ is not a prime:

$$(2m)! = 2m^2 (m^2 -1)(m^2 -4)(m^2 -9) dots (m^2 -(m-1)^2)$$

Notice, $$(2m)!$$ will contain primes of the form $$p_k =m pm n leq 2m$$. The number of such primes are $$pi(2m)$$. Further, the $$m$$ and $$n$$ must be co-prime. Then the number of possible “slots” of the form $$m pm n$$ which can house primes (disregarding $$p_1 = 2$$) are:

$$S(m) = phi(2 m) -1$$

To derive the above the following are considered:

$$1$$. If $$m$$ is even then $$(m^2 – (text{even})^2) = text{even}$$

$$2$$. $$m$$ is not a prime as mentioned before.

$$3$$. $$m – (m-1)$$ is cannot be a prime.

Now, if:

$$pi(2m) > frac{S(m)}{2}$$

Then we can prove Goldbach’s conjecture for that number as that would imply one of the slots $$m+n = p_i$$ has a corresponding slot housing a prime $$m+n =p_k$$. And thus, if add both of them:

$$2m = p_i + p_k$$

## Example

Consider $$m = 15 = 3 cdot 5$$. After considering $$30!$$ eliminating “slots”:

$$S(15) = (1-frac{1}{3})(1-frac{1}{5}) 30 – 1= 15$$

where the remaining slots are:

$$(m^2 -1),(m^2 -7^2),(m^2 -11^2),(m^2 -13^2),(m^2 -17^2),(m^2 -19^2),(m^2 -23^2),(m + 29)$$

However, the number of primes are $$pi(30) = 10$$. Since, there are more primes than “slots” available for them. Then one of the slots of the form $$m$$.

(Note, I didn’t read the background, just the question.)

One has
$$pi(2x) sim frac{2x}{log x}$$
by the prime number theorem and
and
$$phi(x) gg frac{e^{-gamma} x}{log log x}$$
by standard estimates (see the wikipedia page https://en.wikipedia.org/wiki/Euler%27s_totient_function), so since the latter grows much faster there can only be finitely many integers satisfying your inequality (and this finite set has density zero, obviously). More precisely, one has:

$$frac{n}{phi(n)} le e^{gamma} left( log log n + frac{2.5}{e^{gamma} log log n} right)$$
for all $$n ge 3$$ except for $$n = 3 times 5 times 7 times 11 times 13 times 17 times 19 times 23$$,

(See Lemma 4 here: https://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/grytczuk.pdf)

and
$$pi(x) < 1.25506 cdot frac{x}{log x}$$

for $$x ge 17$$ (see https://en.wikipedia.org/wiki/Prime-counting_function) and just from these inequalities (computing $$phi(223092870)$$ by hand) we already get
$$frac{phi(2x) – 1}{2} > pi(2x), quad x > 10^6$$

But then the smaller examples one can check by computer, and find the inequality you want holds for exactly $$649$$ integers, the largest one being $$45045$$ with
$$frac{phi(90090) – 1}{2} = 8639.5 < 8726 = pi(90090).$$

We can rephrase your question as: for which $$m$$ does
$$pi(2m) geq frac{phi(2 m)}{2}$$
hold, or equivalentely
$$frac{phi(2m)}{pi(2m)} leq 2.$$
Now for large enough $$m$$, $$pi(m) approx frac{m}{log(m)}$$, so this inequality becomes (roughly speaking)
$$frac{phi(2m)log(m)}{m} leq 2.$$
One expression for $$phi(k)$$ is $$kprod_{p mid k}left(1-frac1pright)$$, where the variable $$p$$ runs over the primes, so we can rewrite this as
$$log(m)prod_{p mid 2m}left(1-frac1pright) leq 1.$$
Taking logarithms on both sides,
$$log(log(m)) + sum_{p mid 2m}logleft(1-frac1pright) leq 0.$$
For large $$p$$, $$-frac1p$$ is a good approximation for $$logleft(1-frac1pright)$$, so the question becomes
$$log(log(m)) – sum_{p mid 2m} frac1p leq 0.$$
Now a standard approximation tells us that this last sum is approximately equal to $$log(log(p))$$ where $$p$$ is the largest prime in the sum. Clearly the largest prime in this sum will be much smaller than $$m$$, because we took the best case $$m$$: where $$m$$ is a product of lots of distinct small primes.

Thus your inequality should hold almost never.