Diophantine equation 3^a+1=3^b+5^c Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Transforming a Diophantine equation to an elliptic curveNon-negative integer solutions of a single Linear Diophantine EquationDiophantine problemDoes the following Diophantine equation have nontrivial rational solutions?Help with this Diophantine equationHelp with this system of Diophantine equationsThe Theory of Transfinite Diophantine EquationsFind a distinct postive integer solution to this $xyzw=504(x^2+y^2+z^2+w^2)$ diophantine equationExponential diophantine equation systemCombination of $k$-powers and divisibility

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



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Transforming a Diophantine equation to an elliptic curveNon-negative integer solutions of a single Linear Diophantine EquationDiophantine problemDoes the following Diophantine equation have nontrivial rational solutions?Help with this Diophantine equationHelp with this system of Diophantine equationsThe Theory of Transfinite Diophantine EquationsFind a distinct postive integer solution to this $xyzw=504(x^2+y^2+z^2+w^2)$ diophantine equationExponential diophantine equation systemCombination of $k$-powers and divisibility










5












$begingroup$


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
$$
1+3^a = 3^b+5^c.
$$










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Papers are still published on such questions (Lucia gave two examples, and there are others), so it's research-level (and thus fair game for Mathoverflow) even if it did not arise in your own research.
    $endgroup$
    – Noam D. Elkies
    16 mins ago
















5












$begingroup$


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
$$
1+3^a = 3^b+5^c.
$$










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Papers are still published on such questions (Lucia gave two examples, and there are others), so it's research-level (and thus fair game for Mathoverflow) even if it did not arise in your own research.
    $endgroup$
    – Noam D. Elkies
    16 mins ago














5












5








5





$begingroup$


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
$$
1+3^a = 3^b+5^c.
$$










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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









kawakawa

1907




1907







  • 2




    $begingroup$
    Papers are still published on such questions (Lucia gave two examples, and there are others), so it's research-level (and thus fair game for Mathoverflow) even if it did not arise in your own research.
    $endgroup$
    – Noam D. Elkies
    16 mins ago













  • 2




    $begingroup$
    Papers are still published on such questions (Lucia gave two examples, and there are others), so it's research-level (and thus fair game for Mathoverflow) even if it did not arise in your own research.
    $endgroup$
    – Noam D. Elkies
    16 mins ago








2




2




$begingroup$
Papers are still published on such questions (Lucia gave two examples, and there are others), so it's research-level (and thus fair game for Mathoverflow) even if it did not arise in your own research.
$endgroup$
– Noam D. Elkies
16 mins ago





$begingroup$
Papers are still published on such questions (Lucia gave two examples, and there are others), so it's research-level (and thus fair game for Mathoverflow) even if it did not arise in your own research.
$endgroup$
– Noam D. Elkies
16 mins ago











1 Answer
1






active

oldest

votes


















8












$begingroup$

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.



Following a reference from Skinner's paper, Theorem 4.01 of Brenner and Foster gives an explicit treatment of the equation
$$
3^a + 7^b=3^c+5^d,
$$

which completely resolves the problem in this question (take $b=0$). Their proof is elementary, and the only non-trivial solution to the equation in the question is $3^3+1 = 3 + 5^2$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Lucia, many thanks for the paper.
    $endgroup$
    – kawa
    2 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

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.



Following a reference from Skinner's paper, Theorem 4.01 of Brenner and Foster gives an explicit treatment of the equation
$$
3^a + 7^b=3^c+5^d,
$$

which completely resolves the problem in this question (take $b=0$). Their proof is elementary, and the only non-trivial solution to the equation in the question is $3^3+1 = 3 + 5^2$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Lucia, many thanks for the paper.
    $endgroup$
    – kawa
    2 hours ago















8












$begingroup$

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.



Following a reference from Skinner's paper, Theorem 4.01 of Brenner and Foster gives an explicit treatment of the equation
$$
3^a + 7^b=3^c+5^d,
$$

which completely resolves the problem in this question (take $b=0$). Their proof is elementary, and the only non-trivial solution to the equation in the question is $3^3+1 = 3 + 5^2$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Lucia, many thanks for the paper.
    $endgroup$
    – kawa
    2 hours ago













8












8








8





$begingroup$

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.



Following a reference from Skinner's paper, Theorem 4.01 of Brenner and Foster gives an explicit treatment of the equation
$$
3^a + 7^b=3^c+5^d,
$$

which completely resolves the problem in this question (take $b=0$). Their proof is elementary, and the only non-trivial solution to the equation in the question is $3^3+1 = 3 + 5^2$.






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.



Following a reference from Skinner's paper, Theorem 4.01 of Brenner and Foster gives an explicit treatment of the equation
$$
3^a + 7^b=3^c+5^d,
$$

which completely resolves the problem in this question (take $b=0$). Their proof is elementary, and the only non-trivial solution to the equation in the question is $3^3+1 = 3 + 5^2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









LuciaLucia

35k5151177




35k5151177











  • $begingroup$
    Lucia, many thanks for the paper.
    $endgroup$
    – kawa
    2 hours ago
















  • $begingroup$
    Lucia, many thanks for the paper.
    $endgroup$
    – kawa
    2 hours ago















$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
2 hours ago




$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
2 hours ago

















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