Find $7^{2018}-1 bmod 5$ using Fermat's Little Theorem












1












$begingroup$


Let $z=7^{2018}-1$




Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$




$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?




Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
    $endgroup$
    – lulu
    Dec 15 '18 at 13:38






  • 1




    $begingroup$
    Also a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 13:49
















1












$begingroup$


Let $z=7^{2018}-1$




Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$




$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?




Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
    $endgroup$
    – lulu
    Dec 15 '18 at 13:38






  • 1




    $begingroup$
    Also a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 13:49














1












1








1





$begingroup$


Let $z=7^{2018}-1$




Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$




$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?




Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$




Let $z=7^{2018}-1$




Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$




$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?




Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 17:22







Jevaut

















asked Dec 15 '18 at 13:36









JevautJevaut

1,166212




1,166212








  • 2




    $begingroup$
    I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
    $endgroup$
    – lulu
    Dec 15 '18 at 13:38






  • 1




    $begingroup$
    Also a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 13:49














  • 2




    $begingroup$
    I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
    $endgroup$
    – lulu
    Dec 15 '18 at 13:38






  • 1




    $begingroup$
    Also a duplicate of this.
    $endgroup$
    – Jyrki Lahtonen
    Dec 15 '18 at 13:49








2




2




$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38




$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38




1




1




$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49




$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49










3 Answers
3






active

oldest

votes


















1












$begingroup$


Any hints on how to proceed to find zmod5?




Do it the EXACT same way.



$5 -1 = X$



And $2018div X = Q$ with $R$ remaider.



So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.



Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.



.....



But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    We have:
    $$(7^2)^{1009} - 1 pmod 5$$
    $$equiv (-1)^{1009} - 1$$
    $$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @BillDubuque Thanks!
      $endgroup$
      – Toby Mak
      Dec 15 '18 at 14:07










    • $begingroup$
      Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
      $endgroup$
      – Jevaut
      Dec 15 '18 at 15:21












    • $begingroup$
      "Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
      $endgroup$
      – fleablood
      Dec 15 '18 at 17:38



















    1












    $begingroup$

    $7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$



    $7^2=49 ≡-1mod 5$



    $7^{2}.7^{2016}=7^{2018} ≡-1mod 5$



    ⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$


      Any hints on how to proceed to find zmod5?




      Do it the EXACT same way.



      $5 -1 = X$



      And $2018div X = Q$ with $R$ remaider.



      So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.



      Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.



      .....



      But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$


        Any hints on how to proceed to find zmod5?




        Do it the EXACT same way.



        $5 -1 = X$



        And $2018div X = Q$ with $R$ remaider.



        So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.



        Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.



        .....



        But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$


          Any hints on how to proceed to find zmod5?




          Do it the EXACT same way.



          $5 -1 = X$



          And $2018div X = Q$ with $R$ remaider.



          So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.



          Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.



          .....



          But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)






          share|cite|improve this answer









          $endgroup$




          Any hints on how to proceed to find zmod5?




          Do it the EXACT same way.



          $5 -1 = X$



          And $2018div X = Q$ with $R$ remaider.



          So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.



          Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.



          .....



          But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 17:30









          fleabloodfleablood

          72k22687




          72k22687























              2












              $begingroup$

              We have:
              $$(7^2)^{1009} - 1 pmod 5$$
              $$equiv (-1)^{1009} - 1$$
              $$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                @BillDubuque Thanks!
                $endgroup$
                – Toby Mak
                Dec 15 '18 at 14:07










              • $begingroup$
                Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
                $endgroup$
                – Jevaut
                Dec 15 '18 at 15:21












              • $begingroup$
                "Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
                $endgroup$
                – fleablood
                Dec 15 '18 at 17:38
















              2












              $begingroup$

              We have:
              $$(7^2)^{1009} - 1 pmod 5$$
              $$equiv (-1)^{1009} - 1$$
              $$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                @BillDubuque Thanks!
                $endgroup$
                – Toby Mak
                Dec 15 '18 at 14:07










              • $begingroup$
                Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
                $endgroup$
                – Jevaut
                Dec 15 '18 at 15:21












              • $begingroup$
                "Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
                $endgroup$
                – fleablood
                Dec 15 '18 at 17:38














              2












              2








              2





              $begingroup$

              We have:
              $$(7^2)^{1009} - 1 pmod 5$$
              $$equiv (-1)^{1009} - 1$$
              $$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$






              share|cite|improve this answer











              $endgroup$



              We have:
              $$(7^2)^{1009} - 1 pmod 5$$
              $$equiv (-1)^{1009} - 1$$
              $$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 15 '18 at 14:06

























              answered Dec 15 '18 at 13:46









              Toby MakToby Mak

              3,53111128




              3,53111128












              • $begingroup$
                @BillDubuque Thanks!
                $endgroup$
                – Toby Mak
                Dec 15 '18 at 14:07










              • $begingroup$
                Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
                $endgroup$
                – Jevaut
                Dec 15 '18 at 15:21












              • $begingroup$
                "Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
                $endgroup$
                – fleablood
                Dec 15 '18 at 17:38


















              • $begingroup$
                @BillDubuque Thanks!
                $endgroup$
                – Toby Mak
                Dec 15 '18 at 14:07










              • $begingroup$
                Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
                $endgroup$
                – Jevaut
                Dec 15 '18 at 15:21












              • $begingroup$
                "Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
                $endgroup$
                – fleablood
                Dec 15 '18 at 17:38
















              $begingroup$
              @BillDubuque Thanks!
              $endgroup$
              – Toby Mak
              Dec 15 '18 at 14:07




              $begingroup$
              @BillDubuque Thanks!
              $endgroup$
              – Toby Mak
              Dec 15 '18 at 14:07












              $begingroup$
              Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
              $endgroup$
              – Jevaut
              Dec 15 '18 at 15:21






              $begingroup$
              Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
              $endgroup$
              – Jevaut
              Dec 15 '18 at 15:21














              $begingroup$
              "Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
              $endgroup$
              – fleablood
              Dec 15 '18 at 17:38




              $begingroup$
              "Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
              $endgroup$
              – fleablood
              Dec 15 '18 at 17:38











              1












              $begingroup$

              $7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$



              $7^2=49 ≡-1mod 5$



              $7^{2}.7^{2016}=7^{2018} ≡-1mod 5$



              ⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$



                $7^2=49 ≡-1mod 5$



                $7^{2}.7^{2016}=7^{2018} ≡-1mod 5$



                ⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$



                  $7^2=49 ≡-1mod 5$



                  $7^{2}.7^{2016}=7^{2018} ≡-1mod 5$



                  ⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$






                  share|cite|improve this answer









                  $endgroup$



                  $7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$



                  $7^2=49 ≡-1mod 5$



                  $7^{2}.7^{2016}=7^{2018} ≡-1mod 5$



                  ⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 17:59









                  siroussirous

                  1,6851514




                  1,6851514






























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