Show that given any natural number $n$, there are two prime numbers $p$ and $q$ such that $q > p$ and $q - p \geqslant n$ , and all natural numbers strictly between $p$ and $q$ are composite (nonprime) numbers.


closed as off-topic by Martin R, zhoraster, N. F. Taussig, 6005, jameselmore Oct 13 '15 at 15:49

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Take for example $P = n!+1$ and $Q = n! + n+1$, now $Q-P = n$ so $Q>P$.

Now for a number $k$ between $P$ and $Q$ (ie $Q>k>P$) we have $n+1 > k-n! > 1$ so $k = n! + l$ where $n+1 > l > 1$. Now

$$n! = 2\cdot3\cdot...\cdot n$$

Now you can see that since $l\le n$ and $l\ge2$ that $l$ is a factor in the factorial, that is $l|n!$ (ie $n!/l$ is an integer). So

$$k = n! + l = l(n!/l) + l = l(1+n!/l)$$

Finally if $P$ and $Q$ is not primes we could select the largest prime $p\le P$ (exists since $P=n!+1\ge2$) and the smallest prime $q \ge Q$ (exists becaus there exists arbitrarily large primes). Now if k is between $p$ and $q$ it's either between $P$ and $Q$ and it follows that $k$ is composite, or $p < k \le P$ and it's composite because $p$ is the largest prime $\le P$, or $Q \le k < q$ and it's composite because $q$ is the smallest prime $\ge Q$.

  • $\begingroup$ Instead of letting $p=n!+1$, you should let $p$ be the largest prime $\leq n!+1$. Similar comments for $q$. $\endgroup$ – Element118 Oct 12 '15 at 10:48
  • $\begingroup$ @Element118 Yes, but that is not required to prove the statement. One would perhaps need to motivate why there must exist a prime $\le n!+1$ (trivially since $n! \ge 1$). $\endgroup$ – skyking Oct 12 '15 at 11:19
  • $\begingroup$ It is mentioned in the question that p and q are primes, so you need to be a bit more rigorous in writing your proof. $\endgroup$ – Element118 Oct 12 '15 at 11:21
  • $\begingroup$ @Element118 I didn't see that requirement, I've updated the proof to handle that requirement. $\endgroup$ – skyking Oct 12 '15 at 11:58

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