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Let's assume $k$ and $n$ are consecutive prime numbers, $k \lt n$.

An axiom: for any such $k$ and $n$, $k^2 \gt n$.

This seems "obviously" true to me, but could you please prove me wrong? Or if it is correct, could you please help me prove it?

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    $\begingroup$ Since $k \ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate. $\endgroup$ – hardmath Oct 29 '14 at 18:37
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    $\begingroup$ The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$. $\endgroup$ – hardmath Oct 29 '14 at 18:53
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Yes, this is correct, due to Bertrand's Postulate :

Primes occur no further intervals than $n$ and $2n$, and $n^2>2n$ for $n>3$

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    $\begingroup$ Can't figure out why the link is broken, if anyone can fix? $\endgroup$ – Alan Oct 29 '14 at 18:36
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    $\begingroup$ done.${}{}{}{}$ $\endgroup$ – mookid Oct 29 '14 at 18:39

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