本帖最后由 tommywong 于 2018-11-28 18:37 编辑
- $\text{設 } k=v_3(n), n=3^k (3q+r), r=1,2$
- $(2^{3^k}-1)(4^n+2^n+1)\equiv
- (2^{3^k}-1)(4^{3^k(3q+r)}+2^{3^k(3q+r)}+1)\equiv
- (2^{3^k}-1)(4^{3^k r}+2^{3^k r}+1)$
- $\equiv \begin{cases}
- (2^{3^k}-1)(4^{3^k}+2^{3^k}+1)\equiv 2^{3^k 3}-1
- \equiv 0\pmod{8^{3^k}-1}\\
- (2^{3^k}-1)(4^{3^k 2}+2^{3^k 2}+1)
- \equiv 2^{3^k 5}+2^{3^k 3}+2^{3^k}-2^{3^k 4}-2^{3^k 2}-1
- \equiv 0\pmod{8^{3^k}-1}\end{cases}$
- $8^{3^k}-1|(2^{3^k}-1)(4^n+2^n+1)
- \Rightarrow 4^{3^k}+2^{3^k}+1|4^n+2^n+1$
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