# Show that two consecutive integers do not share any prime as factor

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.

• 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 $x+1$? – The Brainlet Exterminator Mar 6 '18 at 6:57

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