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June 24th, 2014, 07:39 AM  #1 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Primes and transcendence?
Here is a coolboy kind of proof for infinitude of primes : $$\prod_p \left ( 1  \frac1{p^2}\right)^{1} = \sum_{k \geq 1} \frac1{k^2} = \frac{\pi^2}6$$ If the leftmost product have finitely many terms then $\pi^2$ would be rational, a contradiction. Now we can use $\zeta(3)$ and it's irrationality to whip up a similar proof. $$\prod_p \left ( 1  \frac1{p^3} \right )^{1} = \sum_{k \geq 1} \frac1{k^3}$$ My question is whether transcendence of $\zeta(3)$ (which is still open) would imply anything stronger than infinitude. Balarka . 
June 24th, 2014, 07:44 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra 
If $p$ in your product denotes "primes", then I think your argument is circular.

June 24th, 2014, 07:48 AM  #3 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 
Yes, by $p$ I mean primes. Why do you think it's circular? AFAICS, neither proof of Basel problem nor the derivation of Euler product involves proving infinitude of primes. Last edited by mathbalarka; June 24th, 2014 at 07:55 AM. 
June 24th, 2014, 08:06 AM  #4 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 
Actually, irrationality (not transcendence!) of $\pi$ (not $\pi^2$!) gives much more than just infinitude of primes : $$\frac{\pi}{4} = \prod_{p = 1(4)} \frac{p}{p1} \cdot \prod_{p = 3(4)} \frac{p}{p+1}$$ So either there are infinitely many primes in the class $1\pmod 4$ or infinitely many in $3\pmod 4$ 
June 24th, 2014, 09:16 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra  Well none of your equalities are true if the sum and product are not infinite! Both involve taking a limit as $p$ or $k$ tend to infinity which so there is an assumption that the sums are infinite..

June 24th, 2014, 09:46 AM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
June 24th, 2014, 11:45 AM  #7  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Quote:
 

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