Is it true...

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• that there exists x,y, a,b 1 s.t. x^a - y^b = n, for all n integers? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/
Message 1 of 7 , Apr 1, 2003
that there exists x,y, a,b>1 s.t. x^a - y^b = n, for all n integers?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• 6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96,. .. (Any pattern there, I wonder?) I asked for a,b 1, hence every odd is accounted
Message 2 of 7 , Apr 2, 2003
'6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96,.
..
(Any pattern there, I wonder?)'

I asked for a,b>1, hence every odd is accounted for, e.g. 6^2-5^2=11,

6,14,32,34,42,46,62,66,70,78,84,86,88,90,96

Is 32 the only 2^k present?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• ... Probably not. I don t think any solution is known for n=6, and I suspect that no solution exists for n=6. For n=6, I think that it can be shown easily
Message 3 of 7 , Apr 2, 2003
> Is it true...
> that there exists x,y, a,b>1 s.t. x^a - y^b = n, for all n integers?

Probably not. I don't think any solution is known for n=6, and I
suspect that no solution exists for n=6. For n=6, I think that it
can be shown easily that if a solution exists, one of the exponents
must be >= 5, and the heuristics (and the scarcity of 5th-and-above
powers) would lead one to strongly suspect that no solution exists.
• In a message dated 03/04/03 02:34:44 GMT Daylight Time, jack@brennen.net ... I agree that heuristics based on density of powers would indicate probably not .
Message 4 of 7 , Apr 3, 2003
In a message dated 03/04/03 02:34:44 GMT Daylight Time, jack@...
writes:

> > Is it true...
> > that there exists x,y, a,b>1 s.t. x^a - y^b = n, for all n integers?
>
> Probably not. I don't think any solution is known for n=6, and I
> suspect that no solution exists for n=6. For n=6, I think that it
> can be shown easily that if a solution exists, one of the exponents
> must be >= 5, and the heuristics (and the scarcity of 5th-and-above
> powers) would lead one to strongly suspect that no solution exists.
>

I agree that heuristics based on density of powers would indicate "probably
not".

Since only 0 and 1 are quadratic residues mod 4, a = b = 2 is no good for n =
2 mod 4.

The next best bet, in terms of density of powers, is a=2, b=3. A couple of
months ago, as a by-product of investigating the "Fermat-Catalan conjecture",
I established that for (x^2) and (y^3) both <= 10^10, no less than 736754 of
the numbers <= 1 million could /not/ be represented as abs(x^2-y^3), the list
starting as follows:-
6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96,...
(Any pattern there, I wonder?)

Mike Oakes

[Non-text portions of this message have been removed]
• In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@globalnet.co.uk ... No. 2048 and 131072 are also in the list. Mike [Non-text portions of this
Message 5 of 7 , Apr 3, 2003
In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@...
writes:

> Is 32 the only 2^k present?
>
>
No.
2048 and 131072 are also in the list.

Mike

[Non-text portions of this message have been removed]
• In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@globalnet.co.uk ... For (x^2) and (y^2) both
Message 6 of 7 , Apr 3, 2003
In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@...
writes:

> I asked for a,b>1, hence every odd is accounted for, e.g. 6^2-5^2=11,
> 7^2-6^2=13, so your list becomes;
>
> 6,14,32,34,42,46,62,66,70,78,84,86,88,90,96
>

For (x^2) and (y^2) both <= 10^10, 19023 (even) integers <= 100000 can't be
expressed as abs(x^2-y^2), the list starting:-
6,14,22,30,38,42,46,54,62,66,70,78,86,94

So 32,34,84,88,90 can be removed from your list, leaving:-
6,14,42,46,62,66,70,78,86,96
needing higher powers.

Except for 96, all these are = 2 mod 4.

Mike Oakes

[Non-text portions of this message have been removed]
• Apologies: I should have written:- ... Of course, 4n = (2n+1)^2 - (2n-1)^2, so only the case n = 2 mod 4 poses any problems. Mike [Non-text portions of this
Message 7 of 7 , Apr 3, 2003
Apologies: I should have written:-

> So 32,34,84,88,90,96 can be removed from your list, leaving:-
> 6,14,42,46,62,66,70,78,86
> needing higher powers.

Of course, 4n = (2n+1)^2 - (2n-1)^2, so only the case n = 2 mod 4 poses any
problems.

Mike

[Non-text portions of this message have been removed]
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