How to calculate all possible values for $m$, where $m=i^k mod p$, $k,p$ are fixed?











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For example, all possible values for $i^{10} mod 71$ is $1, 20, 30, 32, 37, 45, 48$. Is it possible to directly calculate these values without trying all possible $i$ from 1 to 71?






elementary-number-theory modular-arithmetic







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    For example, all possible values for $i^{10} mod 71$ is $1, 20, 30, 32, 37, 45, 48$. Is it possible to directly calculate these values without trying all possible $i$ from 1 to 71?






    elementary-number-theory modular-arithmetic







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      up vote
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      For example, all possible values for $i^{10} mod 71$ is $1, 20, 30, 32, 37, 45, 48$. Is it possible to directly calculate these values without trying all possible $i$ from 1 to 71?






      elementary-number-theory modular-arithmetic







      share|cite|improve this question













      For example, all possible values for $i^{10} mod 71$ is $1, 20, 30, 32, 37, 45, 48$. Is it possible to directly calculate these values without trying all possible $i$ from 1 to 71?






      elementary-number-theory modular-arithmetic




      elementary-number-theory modular-arithmetic






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      asked 2 days ago









      Mayoi

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          In this case $71-1=70$, so as the nonzero residues modulo $71$ form a cyclic
          group of order $70$, the tenth powers form a cyclic group of order $7$.
          So once you have one non-trivial value, say $10$, then $10^0$, $10^1$,
          $10^2,ldots,10^6$ are all of them.






          share|cite|improve this answer





















          • for $i^{374}mod 331$, the full answer is $1, 4, 16, 31, 64, 83, 124, 150, 165, 203, 256, 269, 299, 323, 329$. However, if we take $31$, $64$ to generate the list, only part of the answer can be obtained (of length 3 and 5, while the full answer has 15). Could you elaborate how to deal with this case? Thanks.
            – Mayoi
            2 days ago












          • using the primitive root works, never mind.
            – Mayoi
            2 days ago











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          1 Answer
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          1 Answer
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          up vote
          1
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          accepted










          In this case $71-1=70$, so as the nonzero residues modulo $71$ form a cyclic
          group of order $70$, the tenth powers form a cyclic group of order $7$.
          So once you have one non-trivial value, say $10$, then $10^0$, $10^1$,
          $10^2,ldots,10^6$ are all of them.






          share|cite|improve this answer





















          • for $i^{374}mod 331$, the full answer is $1, 4, 16, 31, 64, 83, 124, 150, 165, 203, 256, 269, 299, 323, 329$. However, if we take $31$, $64$ to generate the list, only part of the answer can be obtained (of length 3 and 5, while the full answer has 15). Could you elaborate how to deal with this case? Thanks.
            – Mayoi
            2 days ago












          • using the primitive root works, never mind.
            – Mayoi
            2 days ago















          up vote
          1
          down vote



          accepted










          In this case $71-1=70$, so as the nonzero residues modulo $71$ form a cyclic
          group of order $70$, the tenth powers form a cyclic group of order $7$.
          So once you have one non-trivial value, say $10$, then $10^0$, $10^1$,
          $10^2,ldots,10^6$ are all of them.






          share|cite|improve this answer





















          • for $i^{374}mod 331$, the full answer is $1, 4, 16, 31, 64, 83, 124, 150, 165, 203, 256, 269, 299, 323, 329$. However, if we take $31$, $64$ to generate the list, only part of the answer can be obtained (of length 3 and 5, while the full answer has 15). Could you elaborate how to deal with this case? Thanks.
            – Mayoi
            2 days ago












          • using the primitive root works, never mind.
            – Mayoi
            2 days ago













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          In this case $71-1=70$, so as the nonzero residues modulo $71$ form a cyclic
          group of order $70$, the tenth powers form a cyclic group of order $7$.
          So once you have one non-trivial value, say $10$, then $10^0$, $10^1$,
          $10^2,ldots,10^6$ are all of them.






          share|cite|improve this answer












          In this case $71-1=70$, so as the nonzero residues modulo $71$ form a cyclic
          group of order $70$, the tenth powers form a cyclic group of order $7$.
          So once you have one non-trivial value, say $10$, then $10^0$, $10^1$,
          $10^2,ldots,10^6$ are all of them.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Lord Shark the Unknown

          96.7k958128




          96.7k958128












          • for $i^{374}mod 331$, the full answer is $1, 4, 16, 31, 64, 83, 124, 150, 165, 203, 256, 269, 299, 323, 329$. However, if we take $31$, $64$ to generate the list, only part of the answer can be obtained (of length 3 and 5, while the full answer has 15). Could you elaborate how to deal with this case? Thanks.
            – Mayoi
            2 days ago












          • using the primitive root works, never mind.
            – Mayoi
            2 days ago


















          • for $i^{374}mod 331$, the full answer is $1, 4, 16, 31, 64, 83, 124, 150, 165, 203, 256, 269, 299, 323, 329$. However, if we take $31$, $64$ to generate the list, only part of the answer can be obtained (of length 3 and 5, while the full answer has 15). Could you elaborate how to deal with this case? Thanks.
            – Mayoi
            2 days ago












          • using the primitive root works, never mind.
            – Mayoi
            2 days ago
















          for $i^{374}mod 331$, the full answer is $1, 4, 16, 31, 64, 83, 124, 150, 165, 203, 256, 269, 299, 323, 329$. However, if we take $31$, $64$ to generate the list, only part of the answer can be obtained (of length 3 and 5, while the full answer has 15). Could you elaborate how to deal with this case? Thanks.
          – Mayoi
          2 days ago






          for $i^{374}mod 331$, the full answer is $1, 4, 16, 31, 64, 83, 124, 150, 165, 203, 256, 269, 299, 323, 329$. However, if we take $31$, $64$ to generate the list, only part of the answer can be obtained (of length 3 and 5, while the full answer has 15). Could you elaborate how to deal with this case? Thanks.
          – Mayoi
          2 days ago














          using the primitive root works, never mind.
          – Mayoi
          2 days ago




          using the primitive root works, never mind.
          – Mayoi
          2 days ago


















           

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