given $p,q,r ge 3$ study the diophantine equation $x^py^q=z^r-1$ using the $abc$-conjecture












1












$begingroup$


I want to show that given $p,q,r ge 3$ the diophantine equation $x^py^q=z^r-1$ has only finitely many solutions with $x,y,z in mathbb{N} = 1 ,2, dots$ assuming the $abc$-conjecture.



The proof supposedly goes in two steps:




  1. $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$

  2. Now apply the $abc$-conjecture to conclude that there are only finitely many solutions.


The statement of the $abc$-conjecture that we use is:



For every $epsilon>0$ there exists only finitely many $ABC$-triples $(a,b,c)in mathbb{N}^3$ s.t. $$c > (text{rad}(abc))^{1+epsilon}.$$



For step 1, I made some observations:




  • $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z$

  • $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z}$


Any help is appreciated; I'm mainly stuck at the first part.



EDIT:



For step 1 we use the further observations $(z^r-1) < z^r$, and $z le z^{frac{r}{3}}$, to find that $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z < z^{frac{2r}{3}}$.



EDIT:
For step 2 take $epsilon = frac{1}{2}$, then the $abc$-conjecture states that there are only finite solutions with $z^r > text{rad}(x^py^qz^r)^{1 + epsilon} = text{rad}(x^py^qz^r)^{1 + frac{1}{2}} = text{rad}(x^py^qz^r)^{frac{3}{2}}$.



But from step 1 we have $z^{frac{2r}{3}} > text{rad}(x^py^qz^r)$ which implies $z^r > text{rad}(x^py^qz^r)^{frac{3}{2}}$, allowing us to conclude that we have only finitely many solutions.



Is my proof correct?






number-theory proof-verification diophantine-equations radicals abc-conjecture







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 18:41










  • $begingroup$
    Do you mean: $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z} le (z^r-1)z le z^r z^{frac{r}{3}}=z^{frac{4r}{3}}$, because that would not work.
    $endgroup$
    – Jens Wagemaker
    Nov 30 '18 at 22:10






  • 1




    $begingroup$
    In your first observation you have $(z^r-1)^{1/3}$, use that.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 22:11
















1












$begingroup$


I want to show that given $p,q,r ge 3$ the diophantine equation $x^py^q=z^r-1$ has only finitely many solutions with $x,y,z in mathbb{N} = 1 ,2, dots$ assuming the $abc$-conjecture.



The proof supposedly goes in two steps:




  1. $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$

  2. Now apply the $abc$-conjecture to conclude that there are only finitely many solutions.


The statement of the $abc$-conjecture that we use is:



For every $epsilon>0$ there exists only finitely many $ABC$-triples $(a,b,c)in mathbb{N}^3$ s.t. $$c > (text{rad}(abc))^{1+epsilon}.$$



For step 1, I made some observations:




  • $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z$

  • $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z}$


Any help is appreciated; I'm mainly stuck at the first part.



EDIT:



For step 1 we use the further observations $(z^r-1) < z^r$, and $z le z^{frac{r}{3}}$, to find that $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z < z^{frac{2r}{3}}$.



EDIT:
For step 2 take $epsilon = frac{1}{2}$, then the $abc$-conjecture states that there are only finite solutions with $z^r > text{rad}(x^py^qz^r)^{1 + epsilon} = text{rad}(x^py^qz^r)^{1 + frac{1}{2}} = text{rad}(x^py^qz^r)^{frac{3}{2}}$.



But from step 1 we have $z^{frac{2r}{3}} > text{rad}(x^py^qz^r)$ which implies $z^r > text{rad}(x^py^qz^r)^{frac{3}{2}}$, allowing us to conclude that we have only finitely many solutions.



Is my proof correct?






number-theory proof-verification diophantine-equations radicals abc-conjecture







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 18:41










  • $begingroup$
    Do you mean: $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z} le (z^r-1)z le z^r z^{frac{r}{3}}=z^{frac{4r}{3}}$, because that would not work.
    $endgroup$
    – Jens Wagemaker
    Nov 30 '18 at 22:10






  • 1




    $begingroup$
    In your first observation you have $(z^r-1)^{1/3}$, use that.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 22:11














1












1








1





$begingroup$


I want to show that given $p,q,r ge 3$ the diophantine equation $x^py^q=z^r-1$ has only finitely many solutions with $x,y,z in mathbb{N} = 1 ,2, dots$ assuming the $abc$-conjecture.



The proof supposedly goes in two steps:




  1. $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$

  2. Now apply the $abc$-conjecture to conclude that there are only finitely many solutions.


The statement of the $abc$-conjecture that we use is:



For every $epsilon>0$ there exists only finitely many $ABC$-triples $(a,b,c)in mathbb{N}^3$ s.t. $$c > (text{rad}(abc))^{1+epsilon}.$$



For step 1, I made some observations:




  • $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z$

  • $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z}$


Any help is appreciated; I'm mainly stuck at the first part.



EDIT:



For step 1 we use the further observations $(z^r-1) < z^r$, and $z le z^{frac{r}{3}}$, to find that $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z < z^{frac{2r}{3}}$.



EDIT:
For step 2 take $epsilon = frac{1}{2}$, then the $abc$-conjecture states that there are only finite solutions with $z^r > text{rad}(x^py^qz^r)^{1 + epsilon} = text{rad}(x^py^qz^r)^{1 + frac{1}{2}} = text{rad}(x^py^qz^r)^{frac{3}{2}}$.



But from step 1 we have $z^{frac{2r}{3}} > text{rad}(x^py^qz^r)$ which implies $z^r > text{rad}(x^py^qz^r)^{frac{3}{2}}$, allowing us to conclude that we have only finitely many solutions.



Is my proof correct?






number-theory proof-verification diophantine-equations radicals abc-conjecture







share|cite|improve this question











$endgroup$




I want to show that given $p,q,r ge 3$ the diophantine equation $x^py^q=z^r-1$ has only finitely many solutions with $x,y,z in mathbb{N} = 1 ,2, dots$ assuming the $abc$-conjecture.



The proof supposedly goes in two steps:




  1. $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$

  2. Now apply the $abc$-conjecture to conclude that there are only finitely many solutions.


The statement of the $abc$-conjecture that we use is:



For every $epsilon>0$ there exists only finitely many $ABC$-triples $(a,b,c)in mathbb{N}^3$ s.t. $$c > (text{rad}(abc))^{1+epsilon}.$$



For step 1, I made some observations:




  • $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z$

  • $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z}$


Any help is appreciated; I'm mainly stuck at the first part.



EDIT:



For step 1 we use the further observations $(z^r-1) < z^r$, and $z le z^{frac{r}{3}}$, to find that $text{rad}(x^py^qz^r) = text{rad}(xyz) le xyz le (z^r-1)^{frac{1}{3}}z < z^{frac{2r}{3}}$.



EDIT:
For step 2 take $epsilon = frac{1}{2}$, then the $abc$-conjecture states that there are only finite solutions with $z^r > text{rad}(x^py^qz^r)^{1 + epsilon} = text{rad}(x^py^qz^r)^{1 + frac{1}{2}} = text{rad}(x^py^qz^r)^{frac{3}{2}}$.



But from step 1 we have $z^{frac{2r}{3}} > text{rad}(x^py^qz^r)$ which implies $z^r > text{rad}(x^py^qz^r)^{frac{3}{2}}$, allowing us to conclude that we have only finitely many solutions.



Is my proof correct?






number-theory proof-verification diophantine-equations radicals abc-conjecture




number-theory proof-verification diophantine-equations radicals abc-conjecture






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 22:30







Jens Wagemaker

















asked Nov 30 '18 at 18:38









Jens WagemakerJens Wagemaker

550311




550311








  • 1




    $begingroup$
    Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 18:41










  • $begingroup$
    Do you mean: $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z} le (z^r-1)z le z^r z^{frac{r}{3}}=z^{frac{4r}{3}}$, because that would not work.
    $endgroup$
    – Jens Wagemaker
    Nov 30 '18 at 22:10






  • 1




    $begingroup$
    In your first observation you have $(z^r-1)^{1/3}$, use that.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 22:11














  • 1




    $begingroup$
    Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 18:41










  • $begingroup$
    Do you mean: $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z} le (z^r-1)z le z^r z^{frac{r}{3}}=z^{frac{4r}{3}}$, because that would not work.
    $endgroup$
    – Jens Wagemaker
    Nov 30 '18 at 22:10






  • 1




    $begingroup$
    In your first observation you have $(z^r-1)^{1/3}$, use that.
    $endgroup$
    – Wojowu
    Nov 30 '18 at 22:11








1




1




$begingroup$
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
$endgroup$
– Wojowu
Nov 30 '18 at 18:41




$begingroup$
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
$endgroup$
– Wojowu
Nov 30 '18 at 18:41












$begingroup$
Do you mean: $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z} le (z^r-1)z le z^r z^{frac{r}{3}}=z^{frac{4r}{3}}$, because that would not work.
$endgroup$
– Jens Wagemaker
Nov 30 '18 at 22:10




$begingroup$
Do you mean: $text{rad}(x^py^qz^r) = text{rad}{(z^r-1)z} le (z^r-1)z le z^r z^{frac{r}{3}}=z^{frac{4r}{3}}$, because that would not work.
$endgroup$
– Jens Wagemaker
Nov 30 '18 at 22:10




1




1




$begingroup$
In your first observation you have $(z^r-1)^{1/3}$, use that.
$endgroup$
– Wojowu
Nov 30 '18 at 22:11




$begingroup$
In your first observation you have $(z^r-1)^{1/3}$, use that.
$endgroup$
– Wojowu
Nov 30 '18 at 22:11










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