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I once found the following problem meant to be solved at high-school level (some olympiad-level exercise, I guess), and I have never been able to prove it using elementary methods. Does anybody know a solution, please?

Show that if $p_n$ is the $n$-th prime (with $p_1 = 2$) then $p_n \leq \frac {n^2 +3n +4} 4$. In a weaker form, show that $p_n = O(n^2)$, but using only elementary methods.

(Later edit: I remembered that the original statement had an "over $4$" denominator, making the inequality tighter.)

Alex M.
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    In order to prove your result by induction if $f(n)=n^2+3n+4$ we have to prove that there is prime between $f(n)$ and $f(n+1)$ which still an unsolved problem in number theory (even for simple functions as $f=x^2$), If we use Bertrand's postulate we can conclude that $p_n\leq 2^n$ which is not what we need to prove, so I think that it would be very interesting if there is a solution at a high-school level – Elaqqad Apr 12 '15 at 19:11
  • An extremely minor point: $n^2 + 3n + 4$ is always even, and $p_1 = 2 < 8$, so the inequality can be strict. – Robert Soupe Apr 12 '15 at 19:27
  • Ah, okay, with that edit, the question becomes much more difficult (and now the question as first posed seems quite easy). – Robert Soupe Apr 12 '15 at 20:15
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    See my answer to a similar question: http://math.stackexchange.com/a/1092552/9911. This only has to be changed very slightly, to conclude that $\pi({n^2 + 3n + 4\over4}) > n$. – doetoe Apr 12 '15 at 20:19
  • @doetoe: I'm not trained in arithmetic; could you elaborate a little bit more on what to do next with $\pi (n)$, please? I don't know how to link it to my problem. Thank you. – Alex M. Apr 12 '15 at 20:22
  • @AlexM.: $p_n \leq \frac {n^2 +3n +4} 4$ means exactly that $\pi(\frac {n^2 +3n +4} 4) > n$. This is easy to prove with the bound $\pi(n) \ge \frac23 \frac{n}{\log n}$ given in the answer which holds for $n\ge3$. – doetoe Apr 12 '15 at 20:32

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See my answer to a similar question: it can be proved by elementary means (as in Zagier's beautiful "The first 50 million prime numbers") that for $n \ge 3$

$$\pi(n) \ge \frac23 \frac{n}{\log n}$$

where $\pi(x)$ is the number of primes below $x$. The statement that $p_n \leq \frac {n^2 +3n +4} 4$ is equivalent to $\pi\left(\frac {n^2 +3n +4} 4\right) \ge n$. For $n = 1,2$ this can be checked directly, for $n\ge3$ it follows from the above inequality.

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