Here is a really neat proof of the infinitude of primes! It's so simple that it makes me say: "Why didn't I think of it?". In fact, it was made just in 2005, by Filip Saidak, and I was really surprised to know that. So here is the proof:

Let \(n > 1\) be a positive integer. Since \(n\) and \(n+1\) are consecutive integers, they must be coprime, and hence the number \(N_2 = n(n + 1)\) must have at least two different prime factors. Similarly, since the integers \(n(n+1)\) and \(n(n+1)+1\) are consecutive, and therefore coprime, the number \(N_3 = n(n + 1)[n(n + 1) + 1]\) must have at least 3 different prime factors. This can be continued indefinitely.

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TopNewestThat's really nice!!!

Darn I have to come up with something like that. :P

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A different proof also exists for the almost same result and with the same concept , apparently.

The proof goes as -

Why didn't I think of it !Log in to reply

Thas' a cool rearrangement o' what we learn in school....

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It seems that the proof shown is for infinitude of composite numbers. I fail to see how it shows that there are infinite number of primes,using this proof??

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if continued indefinitely, the proof is saying that there is a never ending list of composite numbers with a never ending list of unique prime factors. It is different than a proof by contradiction of a largest prime, and is more like a proof of an infinite number of prime factors

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2^prime-1= prime..That means there r infinite prime no.s

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\(2^p-1\) is not always prime for \(p\) being prime. Eg : \(2^{11}-1=2047=23 \times 89\). So, your statement does not actually prove infinite number of primes.

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Euclid proves the same by induction over 2000 years ago

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Isn't it proof by contradiction ?

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dang

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really

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Amazing!

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