Variation of Catalan Conjecture

[e2m2a]

TL;DR Summary

Interested in finding if proofs exist or have been published on a Diophantine equation.

I know that it has been proven that for the expression x^a -y^b = 1, only has this one integer solution, where x = 3, a =2, y =2, b = 3. I am interested in knowing if there is a proof for this expression: 2x^a - y^a =1 in which there are integer solutions for x,a, and y or if no integer solutions exist except for the trivial case where x,y = 1. How would I find a proof for this if it existed?

 

[anuttarasammyak]

a=0

a=1, x=n, y=2n-1 for any integer n

 

[e2m2a]

a = 0 a = 1? Not sure what you mean. Can you please expound on this?

 

[anuttarasammyak]

2x^a - y^a =1

$$2n^0-m^0=1$$
and
$$2n^1-(2n-1)^1=1$$
satisfy it.

 

[PeroK]

$$2n^0-m^0=1$$
and
$$2n^1-(2n-1)^1=1$$
satisfy it.

The OP stated that he didn't want trivial solutions.

 

[anuttarasammyak]

except for the trivial case where x,y = 1.

We have more solutions than he found.
@e2m2a Are you still looking for further more solutions?

 

[e2m2a]

Thanks Anuttarasammyak and Perok for your replies.

 

[e2m2a]

We have more solutions than he found.
@e2m2a Are you still looking for further more solutions?

Yes, I would be very much interested. Please show me the solutions.

 

[anuttarasammyak]

Some easy observations on m and n satisfying
$$2m^a-n^a=1$$
Say a>1, 1<m<n
m and n are coprime to each other, otherwise LHS > 1
n does not have factor 2, otherwise LHS is an even number.
n < 2^(1/a) m otherwise LHS < 0
n^a=m-1(mod m) or let n=k(mod m), k^a=m-1(mod m)

 

[Gaussian97]

Is $$1=50-49=2\cdot25-49=2\cdot 5^2-7^2$$ considered a trivial solution?

 

[anuttarasammyak]

True trivial ones follow after that as
$$2*(-5)^2-7^2=1$$
$$2*5^2-(-7)^2=1$$
$$2*(-5)^2-(-7)^2=1$$

 

[e2m2a]

Is $$1=50-49=2\cdot25-49=2\cdot 5^2-7^2$$ considered a trivial solution?

Interesting. Thanks.

 

[e2m2a]

True trivial ones follow after that as
$$2*(-5)^2-7^2=1$$
$$2*5^2-(-7)^2=1$$
$$2*(-5)^2-(-7)^2=1$$

Thanks.

 

[anuttarasammyak]

I found another solution
$$2*29^2-41^2=1$$

 

[PeroK]

A simple Python program finds the following:

5; 7
29; 41
169; 239
985; 1393
5,741; 8,119
33,461; 47,321
195,025; 275,807
1,136,689; 1,607,521
6,625,109; 9,369,319

We need to look for solutions with $a > 2$.

 

[PeroK]

PS here are solutions with $y^2 - 2x^2 = 1$:

2; 3
12; 17
70; 99
408; 577
2,378; 3,363
13,860; 19,601
80,782; 114,243
470,832; 665,857
2,744,210; 3,880,899

 

[anuttarasammyak]

A simple Python program finds the following:

5; 7
29; 41
169; 239
985; 1393
5,741; 8,119
33,461; 47,321
195,025; 275,807
1,136,689; 1,607,521
6,625,109; 9,369,319

7-5=2
41-29=12
239-169=70
1393-985=408
8119-5741=2378
...

I observe they coincide with first numbers of the lines of

PS here are solutions with y2−2x2=1:

2; 3
12; 17
70; 99
408; 577
2,378; 3,363
13,860; 19,601
80,782; 114,243

In return
3-2=1
17-12=5
99-70=29
577-408=169
...
coincide with the first numbers of upper column except 1 at the top line.

There should be a simple explanation of this coincidence.

 

[mfb]

OEIS has the sequence for the y that have an integer $y^2 - 2x^2 = 1$ solution.
It can be generated using a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.

The ratio between successive x, approaching 3 + sqrt(8), suggested such a relation.

 

[e2m2a]

OEIS has the sequence for the y that have an integer $y^2 - 2x^2 = 1$ solution.
It can be generated using a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.

The ratio between successive x, approaching 3 + sqrt(8), suggested such a relation.

Thanks everyone for your replies and calculations. Totally awesome! We live in an amazing day when computers can grind out solutions for us so quickly. The ancients and pre-modern mathematicians would have loved the tools we have today to test out their ideas. Thanks again.

 

[PeroK]

OEIS has the sequence for the y that have an integer $y^2 - 2x^2 = 1$ solution.
It can be generated using a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.

The ratio between successive x, approaching 3 + sqrt(8), suggested such a relation.

Those are actually the $x$ values where $2x^2 -1 = y^2$.

 

[PeroK]

Thanks everyone for your replies and calculations. Totally awesome! We live in an amazing day when computers can grind out solutions for us so quickly. The ancients and pre-modern mathematicians would have loved the tools we have today to test out their ideas. Thanks again.

Your equation is a special case of Pell's equation:

https://en.wikipedia.org/wiki/Pell's_equation

There are two versions of this:

The general equation $y^2 - nx^2 = 1$ has quite a lot of solutions.

The general equation $nx^2 - y^2 = 1$ has far fewer. Let's focus on this.

I didn't find any solutions (for $x$ up to $10^6$) for $n = 3$ or $n = 4$. But, found these for $n =5$:

$1, 17, 305, 5473, 98209$

This is also on OEIS:

https://oeis.org/search?q=1,+17,+305,5473,98209&sort=&language=english&go=Search

Checking for $n$ up to $19$ I got solutions for:

$n = 10$: $1, 37, 1405, 53353$

$n = 13$: $5, 6485$

$n = 17$: $1, 65, 4289, 283009$

But, if we increase the power to $3$ and look for positive integer solutions to $nx^3 - y^3 = 1$, then I can only find trivial solutions, such as $9\times 1^3 - 2^3 = 1$.

Similarly, I can't find any solutions for the the fourth of fifth power. So, conjecture:

The general equation $nx^a - y^a = 1$ has no non-trivial solutions for $a > 2$ and $n > 0$?

 

[Gaussian97]

Your equation is a special case of Pell's equation:

https://en.wikipedia.org/wiki/Pell's_equation

There are two versions of this:

The general equation $y^2 - nx^2 = 1$ has quite a lot of solutions.

The general equation $nx^2 - y^2 = 1$ has far fewer. Let's focus on this.

I didn't find any solutions (for $x$ up to $10^6$) for $n = 3$ or $n = 4$. But, found these for $n =5$:

$1, 17, 305, 5473, 98209$

This is also on OEIS:

https://oeis.org/search?q=1,+17,+305,5473,98209&sort=&language=english&go=Search

Checking for $n$ up to $19$ I got solutions for:

$n = 10$: $1, 37, 1405, 53353$

$n = 13$: $5, 6485$

$n = 17$: $1, 65, 4289, 283009$

But, if we increase the power to $3$ and look for positive integer solutions to $nx^3 - y^3 = 1$, then I can only find trivial solutions, such as $9\times 1^3 - 2^3 = 1$.

Similarly, I can't find any solutions for the the fourth of fifth power. So, conjecture:

The general equation $nx^a - y^a = 1$ has no non-trivial solutions for $a > 2$ and $n > 0$?

The case $n^{1/a}\in \mathbb{Z}$ should be restricted by Fermat's last theorem, right?

 

[anuttarasammyak]

Some easy observations on m and n satisfying
$$2m^a-n^a=1$$
n < 2^(1/a) m otherwise LHS < 0

for given integer m the candidate of integer n is
$$\lfloor 2^{1/a} m\rfloor$$