Variation of Catalan Conjecture

In summary: This is an example of how your observations can be used to find integer solutions for some values of ##a## and ##m##:1. Let ##a = 4## and ##m = 2##. Then ##2^{1/a} m = \sqrt{2} \times 2 = 2\sqrt{2}##.2. For ##m = 2##, we get ##n = 2\sqrt{2} - 1 \approx 0.828##.3. But, for ##m = 3##, we get ##n = 2\sqrt{2}\times 3 - 1 = 6\sqrt{2} - 1 \approx
  • #1
e2m2a
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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?
 
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  • #2
a=0

a=1, x=n, y=2n-1 for any integer n
 
  • #3
a = 0 a = 1? Not sure what you mean. Can you please expound on this?
 
  • #4
e2m2a said:
2x^a - y^a =1

[tex]2n^0-m^0=1[/tex]
and
[tex]2n^1-(2n-1)^1=1[/tex]
satisfy it.
 
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  • #5
anuttarasammyak said:
[tex]2n^0-m^0=1[/tex]
and
[tex]2n^1-(2n-1)^1=1[/tex]
satisfy it.
The OP stated that he didn't want trivial solutions.
 
  • #6
e2m2a said:
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?
 
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  • #7
Thanks Anuttarasammyak and Perok for your replies.
 
  • #8
anuttarasammyak said:
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.
 
  • #9
Some easy observations on m and n satisfying
[tex]2m^a-n^a=1[/tex]
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)
 
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  • #10
Is $$1=50-49=2\cdot25-49=2\cdot 5^2-7^2$$ considered a trivial solution?
 
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  • #11
True trivial ones follow after that as
[tex]2*(-5)^2-7^2=1[/tex]
[tex]2*5^2-(-7)^2=1[/tex]
[tex]2*(-5)^2-(-7)^2=1[/tex]
 
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  • #12
Gaussian97 said:
Is $$1=50-49=2\cdot25-49=2\cdot 5^2-7^2$$ considered a trivial solution?
Interesting. Thanks.
 
  • #13
anuttarasammyak said:
True trivial ones follow after that as
[tex]2*(-5)^2-7^2=1[/tex]
[tex]2*5^2-(-7)^2=1[/tex]
[tex]2*(-5)^2-(-7)^2=1[/tex]
Thanks.
 
  • #14
I found another solution
[tex]2*29^2-41^2=1[/tex]
 
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  • #15
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##.
 
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  • #16
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
 
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  • #17
PeroK said:
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

PeroK said:
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.
 
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  • #18
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.
 
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  • #19
mfb said:
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.
 
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  • #20
mfb said:
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##.
 
  • #21
e2m2a said:
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##?
 
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  • #22
PeroK said:
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?
 
  • #23
anuttarasammyak said:
Some easy observations on m and n satisfying
[tex]2m^a-n^a=1[/tex]
n < 2^(1/a) m otherwise LHS < 0
for given integer m the candidate of integer n is
[tex]\lfloor 2^{1/a} m\rfloor[/tex]
 
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Related to Variation of Catalan Conjecture

1. What is the Catalan Conjecture?

The Catalan Conjecture, also known as the Catalan's Conjecture or Mihăilescu's Conjecture, is a mathematical conjecture that states that the only solution to the equation xa - yb = 1, where x, y, a, and b are positive integers, is x = 3, y = 2, a = 2, and b = 3. This conjecture has been proven to be true by Preda Mihăilescu in 2002, and it is considered one of the most important problems in number theory.

2. What is the variation of the Catalan Conjecture?

The variation of the Catalan Conjecture is a generalization of the original conjecture that allows for more than one solution to the equation xa - yb = 1. In this variation, the values of x and y can be any positive integers, and the values of a and b can be any positive integers greater than 1. This variation has also been proven to be true by Preda Mihăilescu in 2002.

3. How is the variation of the Catalan Conjecture related to the original conjecture?

The variation of the Catalan Conjecture is a generalization of the original conjecture. This means that the original conjecture is a special case of the variation, where the values of x, y, a, and b are restricted to x = 3, y = 2, a = 2, and b = 3. Therefore, the variation of the Catalan Conjecture is a more general statement that encompasses the original conjecture.

4. What is the significance of the variation of the Catalan Conjecture?

The variation of the Catalan Conjecture is significant because it provides a more general solution to a problem that has been studied for centuries. It also has applications in other areas of mathematics, such as in the study of Diophantine equations. The proof of this variation also helps to strengthen the proof of the original conjecture.

5. Are there any other variations of the Catalan Conjecture?

Yes, there are other variations of the Catalan Conjecture that have been proposed and studied by mathematicians. Some variations allow for negative values of x and y, while others involve different types of equations. However, the variation that has been proven to be true by Preda Mihăilescu in 2002 is the most well-known and widely accepted variation of the Catalan Conjecture.

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