# Variation of Catalan Conjecture

• A
• e2m2a

#### 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?

PeroK

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?

2x^a - y^a =1

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

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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.

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.

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

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

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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.

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

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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
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.

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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.

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
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##.

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
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?

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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