Generalizing Bertrand's Postulate Is it possible that for any integer $y \ge 2$, there exists an integer $x$ such that if an integer $n \ge x$,  then for all integers $z \le n^y$, there exists a prime $p$ such that $z \le p < z+n$  
I am especially interested in a counter example, an argument that it is not possible, or a explanation of what the Prime Number Theorem or some other deep mathematical theorem states about this possibility.
Thanks very much,
-Larry
 A: We can show (using the PNT and using more or less standard notation here) that for sufficiently large $n$ there is an arbitrarily large $k$ such that $$n< p < (n+1/k)n$$
We can prove it in general--the trick is to find a specific $N$ so that if $n>N$ this will work for given k. Nagura proves it for $k = 5$ and other authors have found much higher $k$ and $N.$
Changing the notation of the OP a little we could try to incorporate the claim into the PNT assertion as   
$$n \leq (n/k)^{y}. $$ 
Then $n^{(1/y-1)} =\frac{1}{n^{(1-1/y)}}\leq 1/k$ or $k \leq n^{1-1/y}.$ For concreteness if $y = 2$ initially we have $k \leq \sqrt{n}.$ As $n$ grows larger we would see $k \leq n^{2/3}$ and so on. So even though $k$ can grow it cannot grow too fast. 
As an aside, if we look at $(n,n+n^{1/k})$ there is a theorem that for sufficiently large $n$ there is a prime on $(n,n+n^{0.525})$ due to Baker and a 1996 result of Ch. Jia that there there is a prime on $(n,n+n^{1/20+\epsilon})$ in which $\epsilon$ goes to zero as n gets large. These used analytic and sieve methods.
Taking the second theorem above, this is equivalent to saying that if $n$ is large enough we could take $k \leq n^{19/20}$ and find a prime in the interval. 
So I think this is not a simple consequence of the PNT if it is true. I can't prove or disprove it.   
