"Given $n,t \in \mathbb{N}$ prove that exist $n$ consecutive natural numbers with each one of them having at least one prime factor repeted $t$ times."

Well, at a first glance I couldn't do anything, so I found this:

How do I prove that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite?

And my point is, with the numbers $i+(n+1)!$ with $1<i\leq n+1$, I get $n$ consecutive composite numbers. And if I do $i+((n+1)!)^t$ will I get what I was asked for?

  • $\begingroup$ Here $0 \not \in \mathbb{N}$ $\endgroup$ – Robson Nov 17 '17 at 17:59
  • $\begingroup$ This approach will not work. Just try it for a small case like $n=1, t=2$. You don’t even get one consecutive value with the desired property. $\endgroup$ – Erick Wong Nov 17 '17 at 18:02
  • 2
    $\begingroup$ Use the Chinese Remainder theorem with a suitable set of prime powers $\endgroup$ – Joffan Nov 17 '17 at 18:02

The system

$$ \left\{\begin{array}{rcl}m&\equiv&0\pmod{2^k}\\ m+1&\equiv&0\pmod{3^k}\\ \ldots &\equiv &\ldots \\ m+n-1 &\equiv &0\pmod{p_n^k}\end{array}\right.$$ has a solution $m\leq (2\cdot 3\cdot\ldots\cdot p_n)^k\approx e^{kn}$ which is associated with an interval $[m,m+n-1]$ of integers fulfilling the wanted constraint.


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