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# 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
2 years, 2 months ago

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Assume $$p_n \# > p_{n+1}^2$$ 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 $$n \geq 4$$.

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago