Timeline for Are these inequalities for primes equivalent?

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when toggle format	what		by	license	comment
May 21, 2015 at 23:55	answer	added	The Masked Avenger		timeline score: 2
May 21, 2015 at 22:23	comment	added	The Masked Avenger		It is believed that there are only finitely many gaps g (with adjacent prime p) such that g^2 > p, and the largest of these has p=113. My vote is that L is the same as Q, based on Anthony's analysis and Robert's results.
May 21, 2015 at 22:18	comment	added	Anthony Quas		I should have said $a-b\ge 2$, so that $ab\ge 2p_{n+1}$.
May 21, 2015 at 21:38	comment	added	Robert Israel		There are least two cases of $p_{n+1} < a b$, namely $p_{4} = 7 < 4 \times 2$ and $p_{9} = 23 < 6 \times 4$. Of course these are not counterexamples to the OP: $p_4^2 - p_3 p_5 = -6$ and $p_9^2 - p_8 p_{10} = -22$.
May 21, 2015 at 20:55	comment	added	Anthony Quas		If you write $p_{n+2}=p_{n+1}+a$ and $p_{n+2}=p_{n+1}-b$, the product is $p_{n+1}^2+(a-b)p_{n+1}-ab$. For $L$ to succeed, but $Q$ to fail, you need $a>b$ hence $a-b\ge 1$ and so $ab\ge p_{n+1}$. This is a question about gaps between primes. Terry Tao's blog terrytao.wordpress.com/2014/08/21/… indicates that $a,b\ll p_{n+1}^{.525}$, but that this is considered a very weak upper bound. For a violation of your inequality, you would need 2 consecutive gaps almost as large as the maximum. This can probably be ruled out by some analytic # theory.
May 21, 2015 at 20:10	history	asked	Clark Kimberling	CC BY-SA 3.0