Modulo Arithmetic: Proof of Basic Property

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If a,b,c,k $inmathbbZcap(Ncup0)$ and a$equiv$b(mod c) then prove that a$^kequiv$b$^k$(mod c). I know how to prove it using induction but I wanted to know if there is a method that only uses number theory.




modular-arithmetic induction totient-function multiplicative-function




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    Of course, the quotient $fraca^k-b^ka-b = a^k-1 + a^k-2 b + cdots + b^k-1 = sum_i=0^k-1 a^k-1-i b^i$ would have to involve a recursive definition at some point (if you're going for full formalization), and recursion and induction are nearly the same thing, so...
    – Daniel Schepler
    Aug 27 at 17:40














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down vote

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If a,b,c,k $inmathbbZcap(Ncup0)$ and a$equiv$b(mod c) then prove that a$^kequiv$b$^k$(mod c). I know how to prove it using induction but I wanted to know if there is a method that only uses number theory.




modular-arithmetic induction totient-function multiplicative-function




share|cite|improve this question


















  • 1




    Of course, the quotient $fraca^k-b^ka-b = a^k-1 + a^k-2 b + cdots + b^k-1 = sum_i=0^k-1 a^k-1-i b^i$ would have to involve a recursive definition at some point (if you're going for full formalization), and recursion and induction are nearly the same thing, so...
    – Daniel Schepler
    Aug 27 at 17:40












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
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1





If a,b,c,k $inmathbbZcap(Ncup0)$ and a$equiv$b(mod c) then prove that a$^kequiv$b$^k$(mod c). I know how to prove it using induction but I wanted to know if there is a method that only uses number theory.




modular-arithmetic induction totient-function multiplicative-function




share|cite|improve this question














If a,b,c,k $inmathbbZcap(Ncup0)$ and a$equiv$b(mod c) then prove that a$^kequiv$b$^k$(mod c). I know how to prove it using induction but I wanted to know if there is a method that only uses number theory.





modular-arithmetic induction totient-function multiplicative-function





share|cite|improve this question













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edited Aug 27 at 18:49

























asked Aug 27 at 17:31









Jimmy

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  • 1




    Of course, the quotient $fraca^k-b^ka-b = a^k-1 + a^k-2 b + cdots + b^k-1 = sum_i=0^k-1 a^k-1-i b^i$ would have to involve a recursive definition at some point (if you're going for full formalization), and recursion and induction are nearly the same thing, so...
    – Daniel Schepler
    Aug 27 at 17:40












  • 1




    Of course, the quotient $fraca^k-b^ka-b = a^k-1 + a^k-2 b + cdots + b^k-1 = sum_i=0^k-1 a^k-1-i b^i$ would have to involve a recursive definition at some point (if you're going for full formalization), and recursion and induction are nearly the same thing, so...
    – Daniel Schepler
    Aug 27 at 17:40







1




1




Of course, the quotient $fraca^k-b^ka-b = a^k-1 + a^k-2 b + cdots + b^k-1 = sum_i=0^k-1 a^k-1-i b^i$ would have to involve a recursive definition at some point (if you're going for full formalization), and recursion and induction are nearly the same thing, so...
– Daniel Schepler
Aug 27 at 17:40




Of course, the quotient $fraca^k-b^ka-b = a^k-1 + a^k-2 b + cdots + b^k-1 = sum_i=0^k-1 a^k-1-i b^i$ would have to involve a recursive definition at some point (if you're going for full formalization), and recursion and induction are nearly the same thing, so...
– Daniel Schepler
Aug 27 at 17:40










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If $a equiv b pmodc$, then these two guys are the same element of the ring of integers modulo $c$. Hence the $k$-th power of one is the same as the $k$-th power of the other.



For this you need to prove that the product in the quotient ring is well defined (no induction) and then you can conclude (with induction, if you really want to be that formal) that multiplying $k$ times one is the same as multiplying $k$ times the other one.






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    If $a equiv b pmodc$, then these two guys are the same element of the ring of integers modulo $c$. Hence the $k$-th power of one is the same as the $k$-th power of the other.



    For this you need to prove that the product in the quotient ring is well defined (no induction) and then you can conclude (with induction, if you really want to be that formal) that multiplying $k$ times one is the same as multiplying $k$ times the other one.






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      up vote
      1
      down vote













      If $a equiv b pmodc$, then these two guys are the same element of the ring of integers modulo $c$. Hence the $k$-th power of one is the same as the $k$-th power of the other.



      For this you need to prove that the product in the quotient ring is well defined (no induction) and then you can conclude (with induction, if you really want to be that formal) that multiplying $k$ times one is the same as multiplying $k$ times the other one.






      share|cite|improve this answer






















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        up vote
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        If $a equiv b pmodc$, then these two guys are the same element of the ring of integers modulo $c$. Hence the $k$-th power of one is the same as the $k$-th power of the other.



        For this you need to prove that the product in the quotient ring is well defined (no induction) and then you can conclude (with induction, if you really want to be that formal) that multiplying $k$ times one is the same as multiplying $k$ times the other one.






        share|cite|improve this answer












        If $a equiv b pmodc$, then these two guys are the same element of the ring of integers modulo $c$. Hence the $k$-th power of one is the same as the $k$-th power of the other.



        For this you need to prove that the product in the quotient ring is well defined (no induction) and then you can conclude (with induction, if you really want to be that formal) that multiplying $k$ times one is the same as multiplying $k$ times the other one.







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        answered Aug 27 at 19:21









        Pedro

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