Consider the following problem:
For every $n\in\mathbb{N}$ there exists a series $a_1,...,a_n$ of consecutive numbers s.t. $\forall i. a_i$ is not a power of a prime number (its factorization exists of at least 2 different primes)
Does the following solution hold?
$x\equiv0(2)$
$x\equiv0(3)$
$x+1\equiv0(5)$
$x+1\equiv0(7)$
$...$
$x+(n-1)\equiv0(p_{2n-1})$
$x+(n-1)\equiv0(p_{2n})$
And then construct the series $x,x+1,...,x+(n-1)$
Using the Chinese remainder theorem, there exists a solution. Because each member of the series is congruent to 0 modulu two different prime numbers, it must be a composite of at least 2 different primes.
Thank you in advance.
Well, that's an easy problem. Powers are rare (prime powers even more so), hence there are arbitrarily large gaps between them, which can be proven in a multitude of ways. Your solution is totally OK. Here is another: let's see how many powers are there below some huge $N$. (Let's be generous and consider all of them, not just prime powers.) Apparently, there are no more than $\lfloor\sqrt N\rfloor+\lfloor\sqrt[3]N\rfloor+\lfloor\sqrt[4]N\rfloor+\dots$ of them, which is less than $\log_2N\cdot\sqrt N$, which means that the largest gap between them is bounded from below with ${N\over\log_2N\cdot\sqrt N}={\sqrt N\over\log_2N}$, which can be made arbitrarily large.