Prove that \(\displaystyle {p}_{n}\# >p_{n+1}^2\) for \(n \geq 4\), where \(p_k\) is the \(k\)th prime number and the primorial is defined as \(\displaystyle {p}_{n}\# = \prod_{k=1}^n p_k\).

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TopNewestAssume \(p_n \# > p_{n+1}^2\) for some \(n\); this will be the inductive hypothesis. Then \(p_{n+1} \# > p_{n+1}^3\), and so we seek to prove that \(p_{n+1}^3 \geq p_{n+2}^2\).

It is known that for \(n \geq 0\), \(2p_{n+1} > p_{n+2}\) (alternate form of Bertrand's postulate). As such, cubing both sides produces \(p_{n+1}^3 > \dfrac{p_{n+2}}{8} p_{n+2}^2\), which implies that \(p_{n+1}^3 > p_{n+2}^2\) if \(p_{n+2} > 8\). For \(n \geq 3\), this is obviously true, so \(p_{n+1}^3 > p_{n+2}^2\) (stronger result than required), and thus \(p_{n+1} \# > p_{n+2}^2\).

The hypothesis is verifiably true for \(n = 4\). By induction, because \(p_n \# > p_{n+1}^2 \longrightarrow p_{n+1} \# > p_{n+2}^2\), we have that the hypothesis is true for all \(n \geq 4\). – Jake Lai · 1 year, 3 months ago

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– Mardokay Mosazghi · 1 year, 3 months ago

Clever way to go about it, did not the alternate form.Log in to reply

– Jake Lai · 1 year, 3 months ago

It's an inequality, by the way. An identity is an exact equivalence.Log in to reply

– Kartik Sharma · 1 year, 3 months ago

Brilliant! I am gonna save this solution(I've learned it from Pi Han Goh).Log in to reply