# Generative function for Goldbach's conjecture

I have written a procedure that, given an integer n, it comes out the polynomial with its coefficients are prime and odd $\leq$ n

For example, procedure(5)= $x^{5}+x^{3}+x^{2}$, procedure(10)=$x^{7}+x^{5}+x^{3}+x^{2}$ etc...

We also proove that the equivalent of Goldbach's Conjecture is that every even number $\geq 5$ can be written as the sum of 2 odd number.

Now they ask me to verify Goldbach's conjecture with the help of this procedure for n=1..50, I know it deals with combinatorics, generative function but I don't get the point and we have never study combinatorial..

Tips would be great help !

• Downvoters: please read the question carefully. OP is not claiming to have proven the Goldbach conjecture. – Alex R. May 15 '17 at 21:19

## 1 Answer

Surely you mean that the Goldbach conjecture is that every even number $\geq 5$ can be written as the sum of 2 odd primes. Then to prove the Goldbach conjecture you would need to show that $a_{2k}>0$ for every $k>2$, where:

$$\sum_{k=3}^\infty a_{2k}x^{2k}=\lim_{n\rightarrow\infty} \mbox{procedure}(n)^2,$$

where the limit is in the sense of formal power series rings. This is because $a_{2k}>0$ iff there are two primes $p,q$ such that $p+q=2k$. So to verify Goldbach up to 50, just look at $\mbox{procedure}(50)^2$.

• Thanks! I didn't understand it fully but what do I do when I have procedure(50)^2 ? – Romain B. May 15 '17 at 21:22
• @Haatox: Work out $\mbox{procedure}(8)^2$ by hand, and see if you can verify Goldbach up to $n=8$ with the result. – Alex R. May 15 '17 at 21:23
• Alex, it looks like this : ${x}^{14}+2\,{x}^{12}+3\,{x}^{10}+2\,{x}^{9}+2\,{x}^{8}+2\,{x}^{7}+{x}^{6}+2\,{x}^{5}+{x}^{4}$. What can I conclude ? That all even power exists so it's ok ? – Romain B. May 15 '17 at 21:26
• Yes. Think about how you got each term. It was the result of multiplying $x^p$ with $x^q$ for primes $p,q$ in your generating function. The coefficients count how many ways you can do this. If there's ever an even number $2k$ that doesn't appear as the result of the multiplication, then there aren't any primes $p,q$ such that $p+q=2k$. – Alex R. May 15 '17 at 21:27
• Ok thanks I got it ! – Romain B. May 15 '17 at 21:42