Respuesta :

LammettHash
[tex]3\equiv3\mod{20}[/tex]
[tex]3^2\equiv9\mod{20}[/tex]
[tex]3^3\equiv27\equiv7\mod{20}[/tex]
[tex]3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}[/tex]

[tex]7\equiv7\mod{20}[/tex]
[tex]7^2\equiv49\equiv9\mod{20}[/tex]
[tex]7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}[/tex]
[tex]7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}[/tex]

[tex]9\equiv9\mod{20}[/tex]
[tex]9^2\equiv3^4\equiv1\mod{20}[/tex]

[tex]11\equiv11\mod{20}[/tex]
[tex]11^2\equiv121\equiv1\mod{20}[/tex]

[tex]13\equiv-7\equiv13\mod{20}[/tex]
[tex]13^2\equiv169\equiv9\mod{20}[/tex]
[tex]13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}[/tex]
[tex]13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}[/tex]

[tex]17\equiv-3\equiv17\mod{20}[/tex]
[tex]17^2\equiv(-3)^2\equiv9\mod{20}[/tex]
[tex]17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}[/tex]
[tex]17^4\equiv(-3)^4\equiv81\equiv1\mod{20}[/tex]

[tex]19\equiv-1\equiv19\mod{20}[/tex]
[tex]19^2\equiv19(-1)\equiv-19\equiv1\mod{20}[/tex]

Generally speaking, a number [tex]x[/tex] coprime to [tex]n[/tex] will be a primitive root of [tex]n[/tex] if we have [tex]x^n\equiv x\mod{n}[/tex], or [tex]x^{n-1}\equiv1\mod{n}[/tex]. In other words, if [tex]x[/tex] is of order [tex]n-1[/tex] modulo [tex]n[/tex], then [tex]x[/tex] is a primitive root of [tex]n[/tex].

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.

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