Unverified Commit 982d6bf4 authored by Shuhao Zhang's avatar Shuhao Zhang Committed by GitHub
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Merge pull request #2549 from tml104/master

修改了“求一个原根”的证明的错误
parents 55800a87 e6112c52
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@@ -37,7 +37,7 @@ $$

又有 $t<\varphi(p)$ ,故 $\gcd(t,\varphi(p))\leqslant t<\varphi(p)$ 。

又 $\gcd(t,\varphi(p))\mid(\varphi(p))$ ,故 $\gcd(t,\varphi(p))$ 必至少整除 $a^{\frac{\varphi(p)}{d_{1}}},a^{\frac{\varphi(p)}{d_{2}}},\ldots,a^{\frac{\varphi(p)}{d_{m}}}$ 中的至少一个,设 $\gcd(t,\varphi(p))\mid a^{\frac{\varphi(p)}{d_{i}}}$ ,则 $a^{\frac{\varphi(p)}{d_{i}}}\equiv a^{\gcd(t,\varphi(p))}\equiv 1\pmod{p}$ 。
又 $\gcd(t,\varphi(p))\mid(\varphi(p))$ ,故 $\gcd(t,\varphi(p))$ 必至少整除 ${\frac{\varphi(p)}{d_{1}}},{\frac{\varphi(p)}{d_{2}}},\ldots,{\frac{\varphi(p)}{d_{m}}}$ 中的至少一个,设 $\gcd(t,\varphi(p))\mid {\frac{\varphi(p)}{d_{i}}}$ ,则 $a^{\frac{\varphi(p)}{d_{i}}}\equiv a^{\gcd(t,\varphi(p))}\equiv 1\pmod{p}$ 。

故假设不成立。