Congruence equation for distinct primes $p,q$.












6












$begingroup$



Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$




This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.



Proof:
Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$



Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$



Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
$square$






elementary-number-theory proof-verification prime-numbers alternative-proof







share|cite|improve this question











$endgroup$

















    6












    $begingroup$



    Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$




    This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.



    Proof:
    Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$



    Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$



    Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
    $square$






    elementary-number-theory proof-verification prime-numbers alternative-proof







    share|cite|improve this question











    $endgroup$















      6












      6








      6





      $begingroup$



      Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$




      This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.



      Proof:
      Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$



      Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$



      Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
      $square$






      elementary-number-theory proof-verification prime-numbers alternative-proof







      share|cite|improve this question











      $endgroup$





      Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$




      This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.



      Proof:
      Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$



      Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$



      Since $p+qlt 2qleq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} equiv 1 pmod{pq}.$$
      $square$






      elementary-number-theory proof-verification prime-numbers alternative-proof




      elementary-number-theory proof-verification prime-numbers alternative-proof






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      edited Jan 4 at 21:50









      greedoid

      45.9k1160116




      45.9k1160116










      asked Sep 3 '18 at 9:02









      Moed Pol BolloMoed Pol Bollo

      5231310




      5231310






















          1 Answer
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          $begingroup$

          Your prof is correct.



          Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
          so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
          and thus a conclusion.






          share|cite|improve this answer









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            $begingroup$

            Your prof is correct.



            Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
            so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
            and thus a conclusion.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Your prof is correct.



              Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
              so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
              and thus a conclusion.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Your prof is correct.



                Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
                so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
                and thus a conclusion.






                share|cite|improve this answer









                $endgroup$



                Your prof is correct.



                Or, since $pmid q^{p-1}-1$ and $qmid p^{q-1}-1$ we have $$pqmid (q^{p-1}-1)(p^{q-1}-1)$$
                so $$pq mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq mid p^{q-1}q^{p-1}$$ we have $$pqmid -q^{p-1}-p^{q-1}+1$$
                and thus a conclusion.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 3 '18 at 9:08









                greedoidgreedoid

                45.9k1160116




                45.9k1160116






























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