# Variation of Catalan Conjecture

• A
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?

• PeroK

anuttarasammyak
Gold Member
a=0

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

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

anuttarasammyak
Gold Member
2x^a - y^a =1

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

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• PeroK
PeroK
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Gold Member
2020 Award
$$2n^0-m^0=1$$
and
$$2n^1-(2n-1)^1=1$$
satisfy it.
The OP stated that he didn't want trivial solutions.

anuttarasammyak
Gold Member
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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• PeroK
Thanks Anuttarasammyak and Perok for your replies.

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
Gold Member
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)

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Gaussian97
Homework Helper
Is $$1=50-49=2\cdot25-49=2\cdot 5^2-7^2$$ considered a trivial solution?

• mfb, PeroK and anuttarasammyak
anuttarasammyak
Gold Member
True trivial ones follow after that as
$$2*(-5)^2-7^2=1$$
$$2*5^2-(-7)^2=1$$
$$2*(-5)^2-(-7)^2=1$$

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

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

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PeroK
Homework Helper
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2020 Award
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##.

• mfb and anuttarasammyak
PeroK
Homework Helper
Gold Member
2020 Award
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
anuttarasammyak
Gold Member
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.

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• PeroK
mfb
Mentor
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.

• 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.
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
PeroK
Homework Helper
Gold Member
2020 Award
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
Homework Helper
Gold Member
2020 Award
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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• mfb
Gaussian97
Homework Helper
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
Gold Member
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
$$\lfloor 2^{1/a} m\rfloor$$

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