Prove there exist $a$ such that $ab pmod p = 1$, where $p$ is a prime number and $bin {0,1,2,…p−1}$












1












$begingroup$


How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number



$$ab pmod p = 1$$






number-theory elementary-number-theory







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  • $begingroup$
    en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
    $endgroup$
    – Lord Shark the Unknown
    Oct 21 '18 at 2:08










  • $begingroup$
    @LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
    $endgroup$
    – Cup
    Oct 21 '18 at 2:17










  • $begingroup$
    There is no inverse if $b = 0$.
    $endgroup$
    – JavaMan
    Oct 21 '18 at 2:22
















1












$begingroup$


How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number



$$ab pmod p = 1$$






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
    $endgroup$
    – Lord Shark the Unknown
    Oct 21 '18 at 2:08










  • $begingroup$
    @LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
    $endgroup$
    – Cup
    Oct 21 '18 at 2:17










  • $begingroup$
    There is no inverse if $b = 0$.
    $endgroup$
    – JavaMan
    Oct 21 '18 at 2:22














1












1








1





$begingroup$


How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number



$$ab pmod p = 1$$






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$




How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number



$$ab pmod p = 1$$






number-theory elementary-number-theory




number-theory elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 20:40









rtybase

11.5k31534




11.5k31534










asked Oct 21 '18 at 2:06









CupCup

326




326












  • $begingroup$
    en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
    $endgroup$
    – Lord Shark the Unknown
    Oct 21 '18 at 2:08










  • $begingroup$
    @LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
    $endgroup$
    – Cup
    Oct 21 '18 at 2:17










  • $begingroup$
    There is no inverse if $b = 0$.
    $endgroup$
    – JavaMan
    Oct 21 '18 at 2:22


















  • $begingroup$
    en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
    $endgroup$
    – Lord Shark the Unknown
    Oct 21 '18 at 2:08










  • $begingroup$
    @LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
    $endgroup$
    – Cup
    Oct 21 '18 at 2:17










  • $begingroup$
    There is no inverse if $b = 0$.
    $endgroup$
    – JavaMan
    Oct 21 '18 at 2:22
















$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08




$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08












$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17




$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17












$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22




$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22










3 Answers
3






active

oldest

votes


















0












$begingroup$

Hint: The statement you want to prove is equivalent to finding
$$
ab+kp=1,
$$
for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as




Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$




E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.



The proof of this theorem is left as an exercise.



Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You need to exclude the case $b=0$.
    Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
      $$b^{p-1}equiv 1 pmod{p}$$
      or
      $$bcdot b^{p-2}equiv 1 pmod{p}$$
      and $a = b^{p-2} pmod{p}$.






      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









        0












        $begingroup$

        Hint: The statement you want to prove is equivalent to finding
        $$
        ab+kp=1,
        $$
        for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as




        Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$




        E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.



        The proof of this theorem is left as an exercise.



        Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Hint: The statement you want to prove is equivalent to finding
          $$
          ab+kp=1,
          $$
          for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as




          Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$




          E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.



          The proof of this theorem is left as an exercise.



          Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Hint: The statement you want to prove is equivalent to finding
            $$
            ab+kp=1,
            $$
            for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as




            Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$




            E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.



            The proof of this theorem is left as an exercise.



            Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.






            share|cite|improve this answer









            $endgroup$



            Hint: The statement you want to prove is equivalent to finding
            $$
            ab+kp=1,
            $$
            for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as




            Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$




            E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.



            The proof of this theorem is left as an exercise.



            Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 21 '18 at 2:18









            TreborTrebor

            99815




            99815























                0












                $begingroup$

                You need to exclude the case $b=0$.
                Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  You need to exclude the case $b=0$.
                  Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    You need to exclude the case $b=0$.
                    Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.






                    share|cite|improve this answer









                    $endgroup$



                    You need to exclude the case $b=0$.
                    Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Oct 21 '18 at 3:53









                    Robert IsraelRobert Israel

                    330k23219473




                    330k23219473























                        0












                        $begingroup$

                        Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
                        $$b^{p-1}equiv 1 pmod{p}$$
                        or
                        $$bcdot b^{p-2}equiv 1 pmod{p}$$
                        and $a = b^{p-2} pmod{p}$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
                          $$b^{p-1}equiv 1 pmod{p}$$
                          or
                          $$bcdot b^{p-2}equiv 1 pmod{p}$$
                          and $a = b^{p-2} pmod{p}$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
                            $$b^{p-1}equiv 1 pmod{p}$$
                            or
                            $$bcdot b^{p-2}equiv 1 pmod{p}$$
                            and $a = b^{p-2} pmod{p}$.






                            share|cite|improve this answer









                            $endgroup$



                            Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
                            $$b^{p-1}equiv 1 pmod{p}$$
                            or
                            $$bcdot b^{p-2}equiv 1 pmod{p}$$
                            and $a = b^{p-2} pmod{p}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Oct 21 '18 at 10:00









                            rtybasertybase

                            11.5k31534




                            11.5k31534






























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