从一个式子说起

Last updated on May 8, 2022 pm

质数拆解？

$$3 = (2+1)$$
$$15 = 3\times5 = (2+1)(4+1)$$
$$255 = 3\times5\times17 = (2+1)(4+1)(16+1)$$
$$65535 = 3\times5\times17\times257 = (2+1)(4+1)(16+1)(32+1)$$

$$2^{2^n}-1 = (2^1+1)(2^2+1)\dots(2^{2^{n-1}}+1)$$

Proof. (Using Mathmatical induction)

We suppose that:

1. $n=1$:
$$2^{2^1}-1 = 2^2-1 = 3 = 2^1+1$$
Claim Proofed.
2. Suppose the equation hold when $n=k$, where $k\in\mathbb{N}^\ast$
$$2^{2^n}-1 = (2^1+1)(2^2+1)\dots(2^{2^{n-1}}+1)$$
Thus the equation must hold when $n=k+1$
Thus:
$$2^{2^{k+1}}-1=2^{2k\times2}-1=2^{(2k)\times2}-1=2^{2k^2}-1=(2^{2^k}-1)(2^{2^k}+1)=2^{2^n}-1 = (2^1+1)(2^2+1)\dots(2^{2^{k-1}}+1)(2^{2^k}+1)$$
Which is surely still hold.
Claim Proofed.

Thus, for any $n\in\mathbb{N}^\ast$, the equation would hold.

素数和费马？

a Fermat number, named after Pierre de Fermat, who first studied them, is a positive integer of the form $F_{n}=2^{2^{n}}+1$ where $n$ is a non-negative integer. ——Wikipedia, Fermat number

$$2^{2^n}-1=F_{1}F_{2}\dots F_{n-1}$$

$$2^{2^n}\equiv1{\pmod {F_{1}F_{2}\dots F_{n-1}}}$$

$$2^{2^n}\equiv1{\pmod {F_{n}}}$$

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Author
Matrew File
Posted on
January 13, 2022
Updated on
May 8, 2022