Question

What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?

Answer 1

`3\equiv3\mod{20}`

`3^2\equiv9\mod{20}`

`3^3\equiv27\equiv7\mod{20}`

`3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}`


`7\equiv7\mod{20}`

`7^2\equiv49\equiv9\mod{20}`

`7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}`

`7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}`


`9\equiv9\mod{20}`

`9^2\equiv3^4\equiv1\mod{20}`


`11\equiv11\mod{20}`

`11^2\equiv121\equiv1\mod{20}`


`13\equiv-7\equiv13\mod{20}`

`13^2\equiv169\equiv9\mod{20}`

`13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}`

`13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}`


`17\equiv-3\equiv17\mod{20}`

`17^2\equiv(-3)^2\equiv9\mod{20}`

`17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}`

`17^4\equiv(-3)^4\equiv81\equiv1\mod{20}`


`19\equiv-1\equiv19\mod{20}`

`19^2\equiv19(-1)\equiv-19\equiv1\mod{20}`

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

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