Friday, 29 April 2016

elementary number theory - Is this proof for 1110equiv1pmod100 correct?



Show that 11^{10} \equiv 1\pmod{100}



here, I asked a question about this same problem and all the answers say that I prove this by binomial expansion.



But by doing binomial expansion to show this is the same as doing 11^{10} mod \ 100 in the calculator, no property is being showed to prove this.



But I think this proof might be right.




\gcd(11, 100) = 1



11^{φ(100)} \equiv 1 \pmod{100}



11^{40} \equiv 1 \pmod{100} \ ⇒



11^{10} *11^{10} *11^{10}* 11^{10}\equiv 1\pmod{100}



by Looking everybody mod \ 100, I can say that 11^{10} \equiv 1\pmod{100}



Answer



This doesn't work because the solutions to x^4 \equiv 1 \pmod{100} are x \equiv 1, 7, 43, 49, 51, 57, 93, 99 \pmod{100}. So from (11^{10})^4 \equiv 1 \pmod{100} the best you can say is that 11^{10} \equiv 1, 7, 43, 49, 51, 57, 93, \text{ or } 99 \pmod{100}.


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