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Can assume a smaller subset of positive integers, as the same result would hold for the bigger set.
Let the two positive integers be $x, x+1$, and one would be even and the other odd. So, there cannot be a common prime factor.
The above approach seems incomplete, although definitely there would be no common factors for an even and an odd number. If there were a simple proof that is rigorous too.

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    $\begingroup$ Wouldn't a similar approach work as in the proof of infinite primes? You decompose $x$ as a product of $p_1,\ldots,p_n$, and then $x+1 = p_1\cdots p_n + 1$, so none of the prime factors divide $x+1$? $\endgroup$ – The Brainlet Exterminator Mar 6 '18 at 6:57
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That proof is no proof at all. The number $6$ is even, the number $9$ is odd and they share a common prime factor: $3$.

Here's a proof: if $p$ is prime and $p$ divided both $n$ and $n+1$, then $p$ divides their difference, that is, $p$ divides $1$, which is impossible.

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Hint: Just prove that $(n,n+1)=1.$

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