证明 （5/8）已弃坑

\(\mathcal{No.5:}\)

• 咕咕咕

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\(\mathcal{No.6:}\)

• \(解：\)

  \(\because\ 2^{68}=(2^3)^{18}*2^{14}=8^{18}*2^{14}\)

  \(\therefore\ 2^{68}\equiv8^{18}*2^{14}\ (\mod\ 19)\)

  \(\because\ 8^{18}\equiv1\ (\mod\ 19)\)

  \(\therefore\ 8^{18}*2^{14}\equiv1*2^{14}\equiv2^{14}\ (\mod\ 19)\)

  \(又\because\ 2^{14}\equiv16384\equiv6\ (\mod\ 19)\)

  \(\therefore\ 2^{68}\mod19=6\)

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\(\mathcal{No.7:}\)

• \(证：\)

  \(\ \ \ \ \ \ a^{25}\equiv a\ (\mod\ 30)\)

  \(\Leftrightarrow\ 30|a^{25}-a\)

  \(\therefore\ 即证30|a^{25}-a\)

  \(\because\ a^{25}-a\)

  \(=a(a^{24}-1^{24})\)

  \(=a(a^{12}-1^{12})(a^{12}+1)\)

  \(=a(a^{6}-1^{6})(a^{6}+1)(a^{12}+1)\)

  \(=a(a^{3}-1^{3})(a^{3}+1^{3})(a^{6}+1)(a^{12}+1)\)

  \(=(a-1)a(a+1)(a^{2}-a+1)(a^{2}+a+1)((a^{2})^{3}+1^{3})((a^{3})^{4}+1^{3})\)

  \(=(a-1)a(a+1)(a^{2}-a+1)(a^{2}+1)(a^{2}+a+1)(a^{4}-a^{2}+1)(a^{4}+1)(a^{8}-a^{4}+1)\)

  \(\because\ a、a+1中必有一数为2倍数；a-1、a、a+1中必有一数为3倍数\)

  \(\therefore\ 6|a^{25}-a\)

  \(\therefore\ 即证5|a^{25}-a\)

  \(又\because\ a-1、a、a+1、a^{2}+1中必有一数为5倍数\)

  \(\therefore\ 5|a^{25}-a\)

  \(\because\gcd(5, 6)=1\)

  \(\therefore\ 30|a^{25}-a\)

  \(\therefore\ a^{25}\equiv a\ (\mod\ 30)\)

  \(证毕\)

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\(\mathcal{No.8:}\)

• \(证：\)

  • \(1)\ 当p=2时，q=3\ \Rightarrow\ q>p，得证\)

  • \(2)\ 当p为奇质数时\)

    \(易得2^{p}\equiv 1\ (\mod\ q)\)

    \(\because\ 2^{q-1}\equiv 1\ (\mod\ q)\)

    \(\Rightarrow\ 2^{q-1}\equiv 2^{p}\ (\mod\ q)\)

    \(\therefore\ p为q-1倍数或q-1因数\)

    \(又\because\ q为奇数(2^{p}-1\ mod\ 2=1，q|2^{p}-1)\)

    \(\therefore\ 2|q-1\)

    \(\therefore\ 2p|q-1(2\not|p且p为质数)\ \Rightarrow\ p|\frac{q-1}{2}\ \Rightarrow\ p\leq\frac{q-1}{2}\)

    \(\because\ \frac{q-1}{2} < q\)

    \(\therefore\ p < q\)

    \(证毕\)

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\(\mathcal{No.9:}\)

• \(解：\)

  • \(1)\ 当m=1时，\varphi(m)=1\)

    \(\therefore\ m_{1}=1\)

  • \(2)\ 当m为质数时，可得\varphi(m)=m-1\)

    \(若要使\varphi(m)\mod\ 2=1，则m-1\mod\ 2=1\)

    \(\therefore\ m\mod\ 2=0\Rightarrow m为偶数\)

    \(综上，m_{2}=2\)

  • \(3)\ 当m为合数时：\)

    • \(1.若m有奇素数因子p，则\varphi(m)=\varphi(p)*\cdots=(p-1)*\cdots\)

      \(\because\ p\mod\ 2=1\)
      \(\therefore\ 2|p-1\ \Rightarrow\ 2|\varphi(m)\)

      \(综上，不存在这样的m使\varphi(m)\mod\ 2=1\)

    • \(2.若m=2^{k}(k\geq 2)，则\varphi(m)=\frac{m}{2}\)

      \(\because\ k\geq 2\Rightarrow\ 4|m\Rightarrow\ 2|\frac{m}{2}\)

      \(综上，不存在这样的m使\varphi(m)\mod\ 2=1\)

  \(综上，m_{1}=1，m_{2}=2\)

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\(\mathcal{No.10:}\)

• \(证：\)

  咕咕咕

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\(\mathcal{No.11:}\)

• \(证：\)

  \(前置芝士：\gcd(x,n)=\gcd(n-x,n)\ \ (n>x)\)

  \(再加上\mathcal{No.9}的讨论，可得出小于m且与m互质的数是成对出现的\ (m > 2)\)

  • \(1)\ 当m=2时，\varphi(2)=1，(\sum_{i=1}^{i<m}i\ (\gcd(i, m)=1))=1=\frac{\varphi(2)*2}{2}，证毕\)
  • \(2)\ 当m>2时，可得每对与m互质的数(x，m-x)的平均数为\frac{m}{2}，与m互质的数总个数为\varphi(m)\)
    \(\therefore\ (\sum_{i=1}^{i<m}i\ (\gcd(i, m)=1))=\frac{\varphi(m)*m}{2}，证毕\)

  \(综上,原结论得证\)

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\(\mathcal{No.12:}\)

• 咕咕咕

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