Bonse's Identity! Interesting!

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 \(\displaystyle {p}_{n}\# = \prod_{k=1}^n p_k\).

Note by Kartik Sharma
5 years, 7 months ago

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Comments

Assume pn#>pn+12p_n \# > p_{n+1}^2 for some nn; this will be the inductive hypothesis. Then pn+1#>pn+13p_{n+1} \# > p_{n+1}^3, and so we seek to prove that pn+13≥pn+22p_{n+1}^3 \geq p_{n+2}^2.

It is known that for n≥0n \geq 0, 2pn+1>pn+22p_{n+1} > p_{n+2} (alternate form of Bertrand's postulate). As such, cubing both sides produces pn+13>pn+28pn+22p_{n+1}^3 > \dfrac{p_{n+2}}{8} p_{n+2}^2, which implies that pn+13>pn+22p_{n+1}^3 > p_{n+2}^2 if pn+2>8p_{n+2} > 8. For n≥3n \geq 3, this is obviously true, so pn+13>pn+22p_{n+1}^3 > p_{n+2}^2 (stronger result than required), and thus pn+1#>pn+22p_{n+1} \# > p_{n+2}^2.

The hypothesis is verifiably true for n=4n = 4. By induction, because pn#>pn+12⟶pn+1#>pn+22p_n \# > p_{n+1}^2 \longrightarrow p_{n+1} \# > p_{n+2}^2, we have that the hypothesis is true for all n≥4n \geq 4.

Jake Lai - 5 years, 6 months ago

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It's an inequality, by the way. An identity is an exact equivalence.

Jake Lai - 5 years, 6 months ago

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

Mardokay Mosazghi - 5 years, 6 months ago

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

Kartik Sharma - 5 years, 6 months ago
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