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
$begingroup$
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
asked 1 hour ago
kawakawa
1707
$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
kawakawa
1707
$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
kawakawa
1707
$endgroup$
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
asked 1 hour ago
kawakawa
1707
asked 1 hour ago
kawakawa
1707
asked 1 hour ago
kawakawa
1707
asked 1 hour ago
kawakawa
1707
asked 1 hour ago
kawakawa
1707
1707
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $3^3+1 = 3 + 5^2$.
answered 39 mins ago
LuciaLucia
35k5151177
$endgroup$
$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
35 mins ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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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.
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 $3^3+1 = 3 + 5^2$.
answered 39 mins ago
LuciaLucia
35k5151177
$endgroup$
$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
35 mins ago
add a comment |
$begingroup$
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 $3^3+1 = 3 + 5^2$.
answered 39 mins ago
LuciaLucia
35k5151177
$endgroup$
$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
35 mins ago
add a comment |
$begingroup$
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 $3^3+1 = 3 + 5^2$.
answered 39 mins ago
LuciaLucia
35k5151177
$endgroup$
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 $3^3+1 = 3 + 5^2$.
answered 39 mins ago
LuciaLucia
35k5151177
edited 10 mins ago
answered 39 mins ago
LuciaLucia
35k5151177
answered 39 mins ago
LuciaLucia
35k5151177
answered 39 mins ago
LuciaLucia
35k5151177
35k5151177
$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
35 mins ago
add a comment |
$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
35 mins ago
$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
35 mins ago
$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
35 mins ago
add a comment |
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