Prove the divisibility of $R^k_npm 1$ by 4











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The question is:




Prove that:



For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.




My Try:




  • First we try to find $a$ such that $R_nequiv a pmod 4$.


Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:




$$a equiv -1 pmod 4$$





  • The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$


Any ideas or notes on the prove?



Is there simpler way to prove so?






elementary-number-theory modular-arithmetic divisibility







share|cite|improve this question


















  • 1




    To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
    – F.Carette
    Nov 22 at 13:06










  • @F.Carette, Thanks.
    – Maged Saeed
    Nov 22 at 13:07










  • @F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
    – Bill Dubuque
    Nov 22 at 15:28

















up vote
1
down vote

favorite












The question is:




Prove that:



For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.




My Try:




  • First we try to find $a$ such that $R_nequiv a pmod 4$.


Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:




$$a equiv -1 pmod 4$$





  • The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$


Any ideas or notes on the prove?



Is there simpler way to prove so?






elementary-number-theory modular-arithmetic divisibility







share|cite|improve this question


















  • 1




    To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
    – F.Carette
    Nov 22 at 13:06










  • @F.Carette, Thanks.
    – Maged Saeed
    Nov 22 at 13:07










  • @F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
    – Bill Dubuque
    Nov 22 at 15:28















up vote
1
down vote

favorite









up vote
1
down vote

favorite











The question is:




Prove that:



For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.




My Try:




  • First we try to find $a$ such that $R_nequiv a pmod 4$.


Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:




$$a equiv -1 pmod 4$$





  • The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$


Any ideas or notes on the prove?



Is there simpler way to prove so?






elementary-number-theory modular-arithmetic divisibility







share|cite|improve this question













The question is:




Prove that:



For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.




My Try:




  • First we try to find $a$ such that $R_nequiv a pmod 4$.


Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:




$$a equiv -1 pmod 4$$





  • The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$


Any ideas or notes on the prove?



Is there simpler way to prove so?






elementary-number-theory modular-arithmetic divisibility




elementary-number-theory modular-arithmetic divisibility






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Nov 22 at 12:57









Maged Saeed

725316




725316








  • 1




    To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
    – F.Carette
    Nov 22 at 13:06










  • @F.Carette, Thanks.
    – Maged Saeed
    Nov 22 at 13:07










  • @F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
    – Bill Dubuque
    Nov 22 at 15:28
















  • 1




    To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
    – F.Carette
    Nov 22 at 13:06










  • @F.Carette, Thanks.
    – Maged Saeed
    Nov 22 at 13:07










  • @F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
    – Bill Dubuque
    Nov 22 at 15:28










1




1




To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 at 13:06




To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 at 13:06












@F.Carette, Thanks.
– Maged Saeed
Nov 22 at 13:07




@F.Carette, Thanks.
– Maged Saeed
Nov 22 at 13:07












@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 at 15:28






@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 at 15:28












1 Answer
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Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies



$!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus



$bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$



therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$



Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.






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    Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies



    $!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus



    $bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$



    therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$



    Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies



      $!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus



      $bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$



      therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$



      Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies



        $!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus



        $bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$



        therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$



        Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.






        share|cite|improve this answer














        Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies



        $!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus



        $bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$



        therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$



        Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.







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        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 16:34

























        answered Nov 22 at 16:18









        Bill Dubuque

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