{"id":224458,"date":"2025-06-03T06:27:02","date_gmt":"2025-06-03T06:27:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=224458"},"modified":"2025-06-03T06:27:05","modified_gmt":"2025-06-03T06:27:05","slug":"show-that-7-is-a-primitive-root-of-71","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/03\/show-that-7-is-a-primitive-root-of-71\/","title":{"rendered":"Show that 7 is a primitive root of 71."},"content":{"rendered":"\n

Show that 7 is a primitive root of 71.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Showing that 7 is a Primitive Root Modulo 71<\/h3>\n\n\n\n

To show that 7 is a primitive root modulo 71<\/strong>, we must demonstrate that the smallest positive integer kk such that: 7k\u22611(mod71)7^k \\equiv 1 \\pmod{71}<\/p>\n\n\n\n

is k=\u03d5(71)k = \\phi(71), where \u03d5\\phi is Euler’s totient function. Since 71 is a prime number<\/strong>, we have: \u03d5(71)=71\u22121=70\\phi(71) = 71 – 1 = 70<\/p>\n\n\n\n

Thus, 7 is a primitive root modulo 71 if and only if<\/strong> the smallest kk such that 7k\u22611(mod71)7^k \\equiv 1 \\pmod{71} is k=70k = 70, and no smaller positive integer<\/strong> satisfies this.<\/p>\n\n\n\n

To verify this, we check that: 7d\u22621(mod71)7^d \\not\\equiv 1 \\pmod{71}<\/p>\n\n\n\n

for every proper divisor<\/strong> dd of 70.<\/p>\n\n\n\n

The positive divisors of 70 are: 1,2,5,7,10,14,351, 2, 5, 7, 10, 14, 35<\/p>\n\n\n\n

We’ll compute 7dmod\u2009\u2009717^d \\mod 71 for each:<\/p>\n\n\n\n