# Thread: something new?holds?why?is it possible to prove?

1. ## something new?holds?why?is it possible to prove?

I observed that phi(x)<2pi(x) for all x=6k k: integer greater than zero.
????

2. Take $x=6\cdot p$ where p is a very large prime.

We have $\phi \left( {6p} \right) = 2 \cdot \left( {p - 1} \right)$

Accordint to you, $p-1\leq \pi (6p)$ should hold.

That is $\frac{{\pi \left( {6p} \right)}}
{{p - 1}}\geq 1$
(1)

However $\pi \left( {6p} \right) \approx \frac{{6p}}
{{\log \left( {6p} \right)}} \approx 6\frac{p}
{{\log \left( p \right)}}$
by the prime number theorem.

And therefore the LHS in (1) tends to 0. This means that (1) doesn't hold for large enough $p$

I want also to ask you something that I was thinking and tell me your opinion If you have.
Can we prove that: N(a)>phi(a)/2+w(a)-pi(a) for all even numbers greater
than 10? N(a) gives the number of unordered partitions of type: (x,y) where x,y odd composites coprimes of a: x+y=a, phi(a) is the Euler's totient function, w(a) gives the number of distinct prime divisors of a and pi(a) gives the number of primes less than equal than even a.

thank you

4. ## Clever rabbit...

I don't imagine you'll find an easy proof of that. If you do, you will have proved Goldbach! Here are some numbers >10 that will help you with your line of reasoning: 58 82 86 92 94 106 116. About 40% of numbers<4000 are "candidate" GBC counterexamples, those for whom the RHS of your equation is nonnegative.

5. maybe the answer will not be easy but its important that we know what we should prove in order to prove the conjecture..right?I dont think its impossible to find a good lower bound for those partitions I mentioned before .I think the bast way is to treat these partitions as special diophantine equations and find lower bound for the number of solutions under specific conditions .I dont know another approach at the moment .What do you think?

6. Originally Posted by lakiz
Can we prove that: N(a)>=phi(a)/2+w(a)-pi(a) for all integers greater than 2
This is true.

To prove it, first consider $a=x+y$ without caring about the restriction of them ( $x,y$) being composite. In this case we have $\phi(a)$ such possible ordered pairs.

Now, there are $\pi( a)-\omega(a)$ primes smaller or equal to $a$ that are coprime to $a$, hence we cannot have more than $2\cdot(\pi(a)-\omega(a))$ ordered pairs (1) in which at least one is a prime. Thus the number of ordered partitions is greater or equal to: $\phi(a)-2(\pi(a)-\omega(a))$

But note that the number of ordered partitions must be twice (*) the number of non-ordered partitions thus: $N(a)\geq{\frac{\phi(a)}{2}-\pi(a)+\omega(a)}$

(*) If $x=y$ then $2x=a$ and so $x|a$ and is not coprime to $a$ unless $x=1$ and so $a=2$

If you pay attention to the proof above, you will realise that the difference between the LHS and the RHS is the cardinal of $
\left\{ {\left( {p,q} \right) \in P^2 /p + q = a \wedge p > q} \right\}
$
where $
P
$
is the set of prime numbers, since when we considered (1) we considered that if we have a prime $p$ coprime to $a$ and smaller than it, $a-p$ may not be prime, but if it were, we'd be counting the ordered pairs twice.

So, if you can prove that the inequality is strict for all even numbers greater than>10 you will have proved Goldbach's Conjecture.

7. can you prove that the equality from above holds only for some small evens and for any a even greater than 10 that the inequality holds always? is it possible to prove that while the a increases the difference between the left hand side and the RHS increases too?The equality holds only for some evens up to 10 .It must not be dificult to prove the inequality since this is something true and becomes more true for greater a's.
What do you think?

8. I think you are right!If I prove this then it will be easy to prove the conjecture...But I dont think that is impossible to find one way to prove it .The Goldbach's conjecture can be solved by investigating and expanding that relationship.