Let $pgt 3$ be an prime. If $sum_{k=1}^{p-1} frac{1}{k}=frac{a}{b}$, where $gcd(a,b)=1$. Prove that $pmid a$....












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  • Proof that $sumlimits_i frac{(p-1)!}i$ is divisible by $p$

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Let $pgt 3$ be an prime. Suppose $$sum_{k=1}^{p-1} frac{1}{k}=frac{a}{b}$$ where $gcd(a,b)=1$. Prove that $a$ is divisible by $p$.




Please give me some hint. Sorry for this types of writing. I am not familiar with this.










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marked as duplicate by rtybase, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














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    $begingroup$
    Related: Wolstenholme's Theorem
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    – TheSimpliFire
    Jan 2 at 9:04






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    HINT: working $pmod p$ you have $$sum_{k=1}^{p-1} k^{-1} = sum_{k=1}^{p-1} k equiv frac{p-1}{2} cdot p equiv 0$$ This makes sense because every integer $1 le k le p-1$ has a unique inverse $pmod p$ in that range.
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    – Crostul
    Jan 2 at 9:29












  • $begingroup$
    Please check your hint .....
    $endgroup$
    – Supriyo Banerjee
    Jan 2 at 9:33










  • $begingroup$
    @Crostul you may have to be careful working mod $p$ though:..for example for some positive integer $n$ the equation $2^{-n} equiv_p 1$, but $2^{-n} not = 1$ for any positive integer $n$.
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    – Mike
    Jan 2 at 17:13


















0












$begingroup$



This question already has an answer here:




  • Proof that $sumlimits_i frac{(p-1)!}i$ is divisible by $p$

    2 answers





Let $pgt 3$ be an prime. Suppose $$sum_{k=1}^{p-1} frac{1}{k}=frac{a}{b}$$ where $gcd(a,b)=1$. Prove that $a$ is divisible by $p$.




Please give me some hint. Sorry for this types of writing. I am not familiar with this.










share|cite|improve this question











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marked as duplicate by rtybase, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    Related: Wolstenholme's Theorem
    $endgroup$
    – TheSimpliFire
    Jan 2 at 9:04






  • 3




    $begingroup$
    HINT: working $pmod p$ you have $$sum_{k=1}^{p-1} k^{-1} = sum_{k=1}^{p-1} k equiv frac{p-1}{2} cdot p equiv 0$$ This makes sense because every integer $1 le k le p-1$ has a unique inverse $pmod p$ in that range.
    $endgroup$
    – Crostul
    Jan 2 at 9:29












  • $begingroup$
    Please check your hint .....
    $endgroup$
    – Supriyo Banerjee
    Jan 2 at 9:33










  • $begingroup$
    @Crostul you may have to be careful working mod $p$ though:..for example for some positive integer $n$ the equation $2^{-n} equiv_p 1$, but $2^{-n} not = 1$ for any positive integer $n$.
    $endgroup$
    – Mike
    Jan 2 at 17:13
















0












0








0





$begingroup$



This question already has an answer here:




  • Proof that $sumlimits_i frac{(p-1)!}i$ is divisible by $p$

    2 answers





Let $pgt 3$ be an prime. Suppose $$sum_{k=1}^{p-1} frac{1}{k}=frac{a}{b}$$ where $gcd(a,b)=1$. Prove that $a$ is divisible by $p$.




Please give me some hint. Sorry for this types of writing. I am not familiar with this.










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Proof that $sumlimits_i frac{(p-1)!}i$ is divisible by $p$

    2 answers





Let $pgt 3$ be an prime. Suppose $$sum_{k=1}^{p-1} frac{1}{k}=frac{a}{b}$$ where $gcd(a,b)=1$. Prove that $a$ is divisible by $p$.




Please give me some hint. Sorry for this types of writing. I am not familiar with this.





This question already has an answer here:




  • Proof that $sumlimits_i frac{(p-1)!}i$ is divisible by $p$

    2 answers








elementary-number-theory prime-numbers divisibility






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edited Jan 2 at 21:21









amWhy

192k28225439




192k28225439










asked Jan 2 at 8:31









Supriyo BanerjeeSupriyo Banerjee

1186




1186




marked as duplicate by rtybase, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by rtybase, amWhy, Davide Giraudo, Cesareo, Leucippus Jan 3 at 0:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Related: Wolstenholme's Theorem
    $endgroup$
    – TheSimpliFire
    Jan 2 at 9:04






  • 3




    $begingroup$
    HINT: working $pmod p$ you have $$sum_{k=1}^{p-1} k^{-1} = sum_{k=1}^{p-1} k equiv frac{p-1}{2} cdot p equiv 0$$ This makes sense because every integer $1 le k le p-1$ has a unique inverse $pmod p$ in that range.
    $endgroup$
    – Crostul
    Jan 2 at 9:29












  • $begingroup$
    Please check your hint .....
    $endgroup$
    – Supriyo Banerjee
    Jan 2 at 9:33










  • $begingroup$
    @Crostul you may have to be careful working mod $p$ though:..for example for some positive integer $n$ the equation $2^{-n} equiv_p 1$, but $2^{-n} not = 1$ for any positive integer $n$.
    $endgroup$
    – Mike
    Jan 2 at 17:13
















  • 1




    $begingroup$
    Related: Wolstenholme's Theorem
    $endgroup$
    – TheSimpliFire
    Jan 2 at 9:04






  • 3




    $begingroup$
    HINT: working $pmod p$ you have $$sum_{k=1}^{p-1} k^{-1} = sum_{k=1}^{p-1} k equiv frac{p-1}{2} cdot p equiv 0$$ This makes sense because every integer $1 le k le p-1$ has a unique inverse $pmod p$ in that range.
    $endgroup$
    – Crostul
    Jan 2 at 9:29












  • $begingroup$
    Please check your hint .....
    $endgroup$
    – Supriyo Banerjee
    Jan 2 at 9:33










  • $begingroup$
    @Crostul you may have to be careful working mod $p$ though:..for example for some positive integer $n$ the equation $2^{-n} equiv_p 1$, but $2^{-n} not = 1$ for any positive integer $n$.
    $endgroup$
    – Mike
    Jan 2 at 17:13










1




1




$begingroup$
Related: Wolstenholme's Theorem
$endgroup$
– TheSimpliFire
Jan 2 at 9:04




$begingroup$
Related: Wolstenholme's Theorem
$endgroup$
– TheSimpliFire
Jan 2 at 9:04




3




3




$begingroup$
HINT: working $pmod p$ you have $$sum_{k=1}^{p-1} k^{-1} = sum_{k=1}^{p-1} k equiv frac{p-1}{2} cdot p equiv 0$$ This makes sense because every integer $1 le k le p-1$ has a unique inverse $pmod p$ in that range.
$endgroup$
– Crostul
Jan 2 at 9:29






$begingroup$
HINT: working $pmod p$ you have $$sum_{k=1}^{p-1} k^{-1} = sum_{k=1}^{p-1} k equiv frac{p-1}{2} cdot p equiv 0$$ This makes sense because every integer $1 le k le p-1$ has a unique inverse $pmod p$ in that range.
$endgroup$
– Crostul
Jan 2 at 9:29














$begingroup$
Please check your hint .....
$endgroup$
– Supriyo Banerjee
Jan 2 at 9:33




$begingroup$
Please check your hint .....
$endgroup$
– Supriyo Banerjee
Jan 2 at 9:33












$begingroup$
@Crostul you may have to be careful working mod $p$ though:..for example for some positive integer $n$ the equation $2^{-n} equiv_p 1$, but $2^{-n} not = 1$ for any positive integer $n$.
$endgroup$
– Mike
Jan 2 at 17:13






$begingroup$
@Crostul you may have to be careful working mod $p$ though:..for example for some positive integer $n$ the equation $2^{-n} equiv_p 1$, but $2^{-n} not = 1$ for any positive integer $n$.
$endgroup$
– Mike
Jan 2 at 17:13












2 Answers
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Well



$$sum_{k=1}^{p-1} frac{1}{k} = sum_{k=1}^{p-1} frac{frac{(p-1)!}{k}}{(p-1)!}$$



Now $frac{(p-1)!}{k}$ is an integer for each such $k$, so writing $A=(p-1)!$, we note that $frac{A}{k} equiv_p (A mod p)(k^{-1} mod p)$. Thus



$$sum_{k=1}^{p-1} frac{(p-1)!}{k} doteq sum_{k=1}^{p-1} frac{A}{k} equiv_p sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p).$$



WE then use the fact that each element $k$ has a unqiue inverse to conclude



$$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k.$$



However, one can check that for a prime $p geq 3$ the following holds: $sum_{k=1}^{p-1} (k mod p) equiv_p 0$, concluding



$$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k equiv_p 0,$$



which gives you what you need to show [make sure you see why, it follows because $p$ does not divide $(p-1)!$].






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    For $kleq {p-1over 2}$ let $q_k := {1over k(p-k)}$



    So we have $$({1over 1}+{1over p-1})+({1over 2}+{1over p-2})+...+({1over {p-1over 2}}+{1over {p+1over 2}}) = {aover b}$$



    $$pq_1+pq_2+...+pq_{p-1over 2} = {aover b}$$



    Let $$q_1+q_2+...+q_{p-1over 2} = {cover (p-1)!}$$ for some integer $c$. So we have $$pcdot c cdot b = acdot (p-1)!implies pmid acdot (p-1)!implies pmid a$$






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      Is this not enough answer to your question?
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      – greedoid
      Jan 8 at 18:46


















    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    Well



    $$sum_{k=1}^{p-1} frac{1}{k} = sum_{k=1}^{p-1} frac{frac{(p-1)!}{k}}{(p-1)!}$$



    Now $frac{(p-1)!}{k}$ is an integer for each such $k$, so writing $A=(p-1)!$, we note that $frac{A}{k} equiv_p (A mod p)(k^{-1} mod p)$. Thus



    $$sum_{k=1}^{p-1} frac{(p-1)!}{k} doteq sum_{k=1}^{p-1} frac{A}{k} equiv_p sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p).$$



    WE then use the fact that each element $k$ has a unqiue inverse to conclude



    $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k.$$



    However, one can check that for a prime $p geq 3$ the following holds: $sum_{k=1}^{p-1} (k mod p) equiv_p 0$, concluding



    $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k equiv_p 0,$$



    which gives you what you need to show [make sure you see why, it follows because $p$ does not divide $(p-1)!$].






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Well



      $$sum_{k=1}^{p-1} frac{1}{k} = sum_{k=1}^{p-1} frac{frac{(p-1)!}{k}}{(p-1)!}$$



      Now $frac{(p-1)!}{k}$ is an integer for each such $k$, so writing $A=(p-1)!$, we note that $frac{A}{k} equiv_p (A mod p)(k^{-1} mod p)$. Thus



      $$sum_{k=1}^{p-1} frac{(p-1)!}{k} doteq sum_{k=1}^{p-1} frac{A}{k} equiv_p sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p).$$



      WE then use the fact that each element $k$ has a unqiue inverse to conclude



      $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k.$$



      However, one can check that for a prime $p geq 3$ the following holds: $sum_{k=1}^{p-1} (k mod p) equiv_p 0$, concluding



      $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k equiv_p 0,$$



      which gives you what you need to show [make sure you see why, it follows because $p$ does not divide $(p-1)!$].






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Well



        $$sum_{k=1}^{p-1} frac{1}{k} = sum_{k=1}^{p-1} frac{frac{(p-1)!}{k}}{(p-1)!}$$



        Now $frac{(p-1)!}{k}$ is an integer for each such $k$, so writing $A=(p-1)!$, we note that $frac{A}{k} equiv_p (A mod p)(k^{-1} mod p)$. Thus



        $$sum_{k=1}^{p-1} frac{(p-1)!}{k} doteq sum_{k=1}^{p-1} frac{A}{k} equiv_p sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p).$$



        WE then use the fact that each element $k$ has a unqiue inverse to conclude



        $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k.$$



        However, one can check that for a prime $p geq 3$ the following holds: $sum_{k=1}^{p-1} (k mod p) equiv_p 0$, concluding



        $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k equiv_p 0,$$



        which gives you what you need to show [make sure you see why, it follows because $p$ does not divide $(p-1)!$].






        share|cite|improve this answer











        $endgroup$



        Well



        $$sum_{k=1}^{p-1} frac{1}{k} = sum_{k=1}^{p-1} frac{frac{(p-1)!}{k}}{(p-1)!}$$



        Now $frac{(p-1)!}{k}$ is an integer for each such $k$, so writing $A=(p-1)!$, we note that $frac{A}{k} equiv_p (A mod p)(k^{-1} mod p)$. Thus



        $$sum_{k=1}^{p-1} frac{(p-1)!}{k} doteq sum_{k=1}^{p-1} frac{A}{k} equiv_p sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p).$$



        WE then use the fact that each element $k$ has a unqiue inverse to conclude



        $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k.$$



        However, one can check that for a prime $p geq 3$ the following holds: $sum_{k=1}^{p-1} (k mod p) equiv_p 0$, concluding



        $$sum_{k=1}^{p-1} (A mod p)(k^{-1} mod p) equiv_p sum_{k=1}^{p-1} (A mod p)k equiv_p 0,$$



        which gives you what you need to show [make sure you see why, it follows because $p$ does not divide $(p-1)!$].







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 17:57

























        answered Jan 2 at 17:29









        MikeMike

        3,567411




        3,567411























            0












            $begingroup$

            For $kleq {p-1over 2}$ let $q_k := {1over k(p-k)}$



            So we have $$({1over 1}+{1over p-1})+({1over 2}+{1over p-2})+...+({1over {p-1over 2}}+{1over {p+1over 2}}) = {aover b}$$



            $$pq_1+pq_2+...+pq_{p-1over 2} = {aover b}$$



            Let $$q_1+q_2+...+q_{p-1over 2} = {cover (p-1)!}$$ for some integer $c$. So we have $$pcdot c cdot b = acdot (p-1)!implies pmid acdot (p-1)!implies pmid a$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Is this not enough answer to your question?
              $endgroup$
              – greedoid
              Jan 8 at 18:46
















            0












            $begingroup$

            For $kleq {p-1over 2}$ let $q_k := {1over k(p-k)}$



            So we have $$({1over 1}+{1over p-1})+({1over 2}+{1over p-2})+...+({1over {p-1over 2}}+{1over {p+1over 2}}) = {aover b}$$



            $$pq_1+pq_2+...+pq_{p-1over 2} = {aover b}$$



            Let $$q_1+q_2+...+q_{p-1over 2} = {cover (p-1)!}$$ for some integer $c$. So we have $$pcdot c cdot b = acdot (p-1)!implies pmid acdot (p-1)!implies pmid a$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Is this not enough answer to your question?
              $endgroup$
              – greedoid
              Jan 8 at 18:46














            0












            0








            0





            $begingroup$

            For $kleq {p-1over 2}$ let $q_k := {1over k(p-k)}$



            So we have $$({1over 1}+{1over p-1})+({1over 2}+{1over p-2})+...+({1over {p-1over 2}}+{1over {p+1over 2}}) = {aover b}$$



            $$pq_1+pq_2+...+pq_{p-1over 2} = {aover b}$$



            Let $$q_1+q_2+...+q_{p-1over 2} = {cover (p-1)!}$$ for some integer $c$. So we have $$pcdot c cdot b = acdot (p-1)!implies pmid acdot (p-1)!implies pmid a$$






            share|cite|improve this answer









            $endgroup$



            For $kleq {p-1over 2}$ let $q_k := {1over k(p-k)}$



            So we have $$({1over 1}+{1over p-1})+({1over 2}+{1over p-2})+...+({1over {p-1over 2}}+{1over {p+1over 2}}) = {aover b}$$



            $$pq_1+pq_2+...+pq_{p-1over 2} = {aover b}$$



            Let $$q_1+q_2+...+q_{p-1over 2} = {cover (p-1)!}$$ for some integer $c$. So we have $$pcdot c cdot b = acdot (p-1)!implies pmid acdot (p-1)!implies pmid a$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 18:10









            greedoidgreedoid

            39.5k114797




            39.5k114797












            • $begingroup$
              Is this not enough answer to your question?
              $endgroup$
              – greedoid
              Jan 8 at 18:46


















            • $begingroup$
              Is this not enough answer to your question?
              $endgroup$
              – greedoid
              Jan 8 at 18:46
















            $begingroup$
            Is this not enough answer to your question?
            $endgroup$
            – greedoid
            Jan 8 at 18:46




            $begingroup$
            Is this not enough answer to your question?
            $endgroup$
            – greedoid
            Jan 8 at 18:46



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