Diophantine equation 3^a+1=3^b+5^c












1












$begingroup$


This is not a research problem, but challenging enough that I've decided to post it in here:



Determine all triples $(a,b,c)$ of non-negative integers, satisfying
$$
1+3^a = 3^b+5^c.
$$






nt.number-theory diophantine-equations elementary-proofs







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    This is not a research problem, but challenging enough that I've decided to post it in here:



    Determine all triples $(a,b,c)$ of non-negative integers, satisfying
    $$
    1+3^a = 3^b+5^c.
    $$






    nt.number-theory diophantine-equations elementary-proofs







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      This is not a research problem, but challenging enough that I've decided to post it in here:



      Determine all triples $(a,b,c)$ of non-negative integers, satisfying
      $$
      1+3^a = 3^b+5^c.
      $$






      nt.number-theory diophantine-equations elementary-proofs







      share|cite|improve this question









      $endgroup$




      This is not a research problem, but challenging enough that I've decided to post it in here:



      Determine all triples $(a,b,c)$ of non-negative integers, satisfying
      $$
      1+3^a = 3^b+5^c.
      $$






      nt.number-theory diophantine-equations elementary-proofs




      nt.number-theory diophantine-equations elementary-proofs






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 43 mins ago









      kawakawa

      1707




      1707






















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          $begingroup$

          I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation
          $$
          ap^x bq^y = c+ dp^z q^w
          $$
          has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in principle) the more general equation $1+3^a 5^d = 3^b+ 5^c$. Anyway, there is a large literature around such exponential diophantine equations, and Skinner's paper will give some references.






          share|cite|improve this answer









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          • $begingroup$
            Lucia, many thanks for the paper.
            $endgroup$
            – kawa
            17 mins ago












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          active

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          active

          oldest

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          4












          $begingroup$

          I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation
          $$
          ap^x bq^y = c+ dp^z q^w
          $$
          has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in principle) the more general equation $1+3^a 5^d = 3^b+ 5^c$. Anyway, there is a large literature around such exponential diophantine equations, and Skinner's paper will give some references.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Lucia, many thanks for the paper.
            $endgroup$
            – kawa
            17 mins ago
















          4












          $begingroup$

          I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation
          $$
          ap^x bq^y = c+ dp^z q^w
          $$
          has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in principle) the more general equation $1+3^a 5^d = 3^b+ 5^c$. Anyway, there is a large literature around such exponential diophantine equations, and Skinner's paper will give some references.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Lucia, many thanks for the paper.
            $endgroup$
            – kawa
            17 mins ago














          4












          4








          4





          $begingroup$

          I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation
          $$
          ap^x bq^y = c+ dp^z q^w
          $$
          has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in principle) the more general equation $1+3^a 5^d = 3^b+ 5^c$. Anyway, there is a large literature around such exponential diophantine equations, and Skinner's paper will give some references.






          share|cite|improve this answer









          $endgroup$



          I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation
          $$
          ap^x bq^y = c+ dp^z q^w
          $$
          has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in principle) the more general equation $1+3^a 5^d = 3^b+ 5^c$. Anyway, there is a large literature around such exponential diophantine equations, and Skinner's paper will give some references.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 21 mins ago









          LuciaLucia

          34.9k5151177




          34.9k5151177












          • $begingroup$
            Lucia, many thanks for the paper.
            $endgroup$
            – kawa
            17 mins ago


















          • $begingroup$
            Lucia, many thanks for the paper.
            $endgroup$
            – kawa
            17 mins ago
















          $begingroup$
          Lucia, many thanks for the paper.
          $endgroup$
          – kawa
          17 mins ago




          $begingroup$
          Lucia, many thanks for the paper.
          $endgroup$
          – kawa
          17 mins ago


















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