本帖最后由 hbghlyj 于 2020-1-16 10:49 编辑
相关问题
设整数n＞1,n|$s_n=2^{\phi(n)}+3^{\phi(n)}+\cdots+n^{\phi(n)}$,记$p_1,p_2\cdots p_k$k为n的全体不同素因子,求证$\frac1{p_1}+\frac1{p_2}+\cdots+\frac1{p_k}+\frac1{p_1p_2\cdots p_k}$是整数- In[3]:= Select[Range[1000],Divisible[Sum[i^EulerPhi[#],{i,2,#}],#]&]
- Out[3]= {1,2,6,42}
证明：先证明n无平方因子.若存在素数p使$n=p^2m$,m,则$1+s_n\equiv0^{\phi(n)}+1^{\phi(n)}+2^{\phi(n)}+3^{\phi(n)}+\cdots+(n-1)^{\phi(n)}\equiv\sum\limits_{j=0}^{mp-1}\sum\limits_{k=0}^{p-1}(jp+k)^{\phi(n)}\equiv\sum\limits_{j=0}^{mp-1}\sum\limits_{k=0}^{p-1}k^{\phi(n)}=mp\sum\limits_{k=0}^{p-1}k^{\phi(n)}\equiv0\pmod p$,即$p|1+s_n$,但$p|n,n|s_n$,矛盾
所以$n=p_1p_2\cdots p_k,p_1＜p_2＜\cdots＜p_k$为素数.因为$\phi(p_i)|\phi(n)$,由费马小定理$s_n\equiv n-\frac n{p_i}-1\equiv0\pmod p_i$,所以$p_i|1+\frac n{p_i}$,i,$p_1p_2\cdots p_k|\frac n{p_1}+\frac n{p_2}+\cdots+\frac n{p_k}n+1=p_2p_3\cdots p_k+p_1p_3\cdots p_k+\cdots+p_1p_2\cdots p_{k-1}+1$ |
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发表于 2020-1-9 13:20
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