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












10












$begingroup$


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











$endgroup$








  • 3




    $begingroup$
    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
    Apr 22 at 2:47








  • 2




    $begingroup$
    But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO.
    $endgroup$
    – Gerry Myerson
    Apr 22 at 12:35










  • $begingroup$
    NoamD.Elkies: thanks for your comment. GerryMyerson: thanks. Had I known that these type of Diophantine equations are still an active area of research, would have likely phrased the problem that way. Still trying to digest the concept --- as there is a section (tag) with elementary proofs, that are contest problems, as opposed to research problems, but challenging enough that people still post.
    $endgroup$
    – kawa
    Apr 22 at 14:14


















10












$begingroup$


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











$endgroup$








  • 3




    $begingroup$
    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
    Apr 22 at 2:47








  • 2




    $begingroup$
    But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO.
    $endgroup$
    – Gerry Myerson
    Apr 22 at 12:35










  • $begingroup$
    NoamD.Elkies: thanks for your comment. GerryMyerson: thanks. Had I known that these type of Diophantine equations are still an active area of research, would have likely phrased the problem that way. Still trying to digest the concept --- as there is a section (tag) with elementary proofs, that are contest problems, as opposed to research problems, but challenging enough that people still post.
    $endgroup$
    – kawa
    Apr 22 at 14:14
















10












10








10


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











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








edited Apr 22 at 5:31









TheSimpliFire

12510




12510










asked Apr 22 at 0:23









kawakawa

2188




2188








  • 3




    $begingroup$
    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
    Apr 22 at 2:47








  • 2




    $begingroup$
    But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO.
    $endgroup$
    – Gerry Myerson
    Apr 22 at 12:35










  • $begingroup$
    NoamD.Elkies: thanks for your comment. GerryMyerson: thanks. Had I known that these type of Diophantine equations are still an active area of research, would have likely phrased the problem that way. Still trying to digest the concept --- as there is a section (tag) with elementary proofs, that are contest problems, as opposed to research problems, but challenging enough that people still post.
    $endgroup$
    – kawa
    Apr 22 at 14:14
















  • 3




    $begingroup$
    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
    Apr 22 at 2:47








  • 2




    $begingroup$
    But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO.
    $endgroup$
    – Gerry Myerson
    Apr 22 at 12:35










  • $begingroup$
    NoamD.Elkies: thanks for your comment. GerryMyerson: thanks. Had I known that these type of Diophantine equations are still an active area of research, would have likely phrased the problem that way. Still trying to digest the concept --- as there is a section (tag) with elementary proofs, that are contest problems, as opposed to research problems, but challenging enough that people still post.
    $endgroup$
    – kawa
    Apr 22 at 14:14










3




3




$begingroup$
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
Apr 22 at 2:47






$begingroup$
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
Apr 22 at 2:47






2




2




$begingroup$
But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO.
$endgroup$
– Gerry Myerson
Apr 22 at 12:35




$begingroup$
But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO.
$endgroup$
– Gerry Myerson
Apr 22 at 12:35












$begingroup$
NoamD.Elkies: thanks for your comment. GerryMyerson: thanks. Had I known that these type of Diophantine equations are still an active area of research, would have likely phrased the problem that way. Still trying to digest the concept --- as there is a section (tag) with elementary proofs, that are contest problems, as opposed to research problems, but challenging enough that people still post.
$endgroup$
– kawa
Apr 22 at 14:14






$begingroup$
NoamD.Elkies: thanks for your comment. GerryMyerson: thanks. Had I known that these type of Diophantine equations are still an active area of research, would have likely phrased the problem that way. Still trying to digest the concept --- as there is a section (tag) with elementary proofs, that are contest problems, as opposed to research problems, but challenging enough that people still post.
$endgroup$
– kawa
Apr 22 at 14:14












1 Answer
1






active

oldest

votes


















25












$begingroup$

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
    Apr 22 at 0:48












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









25












$begingroup$

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
    Apr 22 at 0:48
















25












$begingroup$

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
    Apr 22 at 0:48














25












25








25





$begingroup$

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 Apr 22 at 1:13

























answered Apr 22 at 0:44









LuciaLucia

35.2k5152179




35.2k5152179












  • $begingroup$
    Lucia, many thanks for the paper.
    $endgroup$
    – kawa
    Apr 22 at 0:48


















  • $begingroup$
    Lucia, many thanks for the paper.
    $endgroup$
    – kawa
    Apr 22 at 0:48
















$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
Apr 22 at 0:48




$begingroup$
Lucia, many thanks for the paper.
$endgroup$
– kawa
Apr 22 at 0:48


















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