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

User Misty
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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.
User Steve Hatcher
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8.2k points

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