given $p,q,r ge 3$ study the diophantine equation $x^py^q=z^r-1$ using the $abc$-conjecture
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:
- $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$
- 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
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
add a comment |
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:
- $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$
- 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
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
1
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
– Wojowu
Nov 30 '18 at 18:41
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.
– Jens Wagemaker
Nov 30 '18 at 22:10
1
In your first observation you have $(z^r-1)^{1/3}$, use that.
– Wojowu
Nov 30 '18 at 22:11
add a comment |
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:
- $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$
- 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
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
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:
- $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$
- 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
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
edited Nov 30 '18 at 22:30
Jens Wagemaker
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
asked Nov 30 '18 at 18:38
Jens WagemakerJens Wagemaker
550311
550311
1
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
– Wojowu
Nov 30 '18 at 18:41
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.
– Jens Wagemaker
Nov 30 '18 at 22:10
1
In your first observation you have $(z^r-1)^{1/3}$, use that.
– Wojowu
Nov 30 '18 at 22:11
add a comment |
1
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
– Wojowu
Nov 30 '18 at 18:41
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.
– Jens Wagemaker
Nov 30 '18 at 22:10
1
In your first observation you have $(z^r-1)^{1/3}$, use that.
– Wojowu
Nov 30 '18 at 22:11
1
1
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
– Wojowu
Nov 30 '18 at 18:41
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
– Wojowu
Nov 30 '18 at 18:41
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.
– Jens Wagemaker
Nov 30 '18 at 22:10
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.
– Jens Wagemaker
Nov 30 '18 at 22:10
1
1
In your first observation you have $(z^r-1)^{1/3}$, use that.
– Wojowu
Nov 30 '18 at 22:11
In your first observation you have $(z^r-1)^{1/3}$, use that.
– Wojowu
Nov 30 '18 at 22:11
add a comment |
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1
Hint for first part: $z^r-1<z^r,zleq z^{r/3}$.
– Wojowu
Nov 30 '18 at 18:41
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.
– Jens Wagemaker
Nov 30 '18 at 22:10
1
In your first observation you have $(z^r-1)^{1/3}$, use that.
– Wojowu
Nov 30 '18 at 22:11