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June 24th, 2014, 08:39 AM   #1
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Primes and transcendence?

Here is a cool-boy 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
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June 24th, 2014, 08:44 AM   #2
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If $p$ in your product denotes "primes", then I think your argument is circular.
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June 24th, 2014, 08:48 AM   #3
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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 08:55 AM.
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June 24th, 2014, 09:06 AM   #4
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Actually, irrationality (not transcendence!) of $\pi$ (not $\pi^2$!) gives much more than just infinitude of primes :

$$\frac{\pi}{4} = \prod_{p = 1(4)} \frac{p}{p-1} \cdot \prod_{p = 3(4)} \frac{p}{p+1}$$

So either there are infinitely many primes in the class $1\pmod 4$ or infinitely many in $3\pmod 4$
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June 24th, 2014, 10:16 AM   #5
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Originally Posted by mathbalarka View Post
AFAICS, neither proof of Basel problem nor the derivation of Euler product involves proving infinitude of primes.
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..
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June 24th, 2014, 10:46 AM   #6
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Well none of your equalities are true if the sum and product are not infinite!
...and this is precisely the proof.
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June 24th, 2014, 12:45 PM   #7
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Well none of your equalities are true if the sum and product are not infinite!
That is the proof, and there is nothing circular to it.
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