I think I proved something, but to be sure I'm right about it, I'm gonna let others trying to prove it (or disprove it) themselves. I'm not so good in all that math symbol thing, so I'm just going to pick a random symbol for all the things here. just go with it please.

let P(n) be the nth prime. for example P(1) = 2, P(2) = 3 , P(4) = 7....

now let Φ(n) = P(1) * P(2) * P(3) ... * P(n)

so what I think I proved is that both Φ(n)+1 and Φ(n) - 1 are primes (twin primes actually).

it's a simple proof, but I think I might be wrong here so I invite anyone who wants to, to prove or disprove it.

hint: you can easily prove that Φ(n) is not divisible by any prime numbers under P(n), but what about prime numbers that are bigger than P(n)?

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## Comments

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TopNewestWell, \(\phi(4)-1=209=11\cdot 19\).

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oh well here goes my theory. I hoped it was true, but I guess I knew it had to be wrong. thanks for the counterexample. it actually explains to me why my "proof" was wrong.

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this is a counterexample

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This is not the first time I've seen this.

To be honest, this is the first thing that came to my mind too when I was introduced to the Twin Prime Problem for the very first time.

It turns out that even \(\phi(n)+1\) can be composite.

Take \(\phi(6)+1=2\times3\times5\times7\times11\times13+1=30031=59\times509\).

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really? I just thought about it for the first time yesterday when I went for a walk, years after I first heard about twin primes. dammit the people on this site are too smart for me.

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Yes, there are some really smart people on this site!

While you proved that all primes up to \(P(n)\) don't divide \(\phi(n) \pm 1\) [I'm using your notation], you didn't prove that primes larger than \(P(n)\) don't divide \(\phi(n) \pm 1\).

As \(n\) gets larger, \(\phi(n) \pm 1\) gets really huge leaving a lot of room for other primes. But I think you now know this.

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