|2^n-3^m|=1

[a1call]

|2^n-3^m|=1

|2^1-3^1|= |2^2-3^1|= |2^3-3^2|=1

Could there be any other m and n combo for which |2^n-3^m|=1 over the integer domain?

I think it is not absolutely impossible (though it is virtually impossible).
Is it?
Can it be proven?

[paulunderwood]

https://en.wikipedia.org/wiki/Catalan%27s_conjecture

[a1call]

Originally Posted by paulunderwood

https://en.wikipedia.org/wiki/Catalan%27s_conjecture

Thank you very much for that link paulunderwood,
It should save me from a few sleepless nights.

Relatively, this should be a good use of one's CPU's idle clocks:
https://en.m.wikipedia.org/wiki/Tijd...njecture#Prize

[science_man_88]

Originally Posted by a1call

Thank you very much for that link paulunderwood,
It should save me from a few sleepless nights.

Relatively, this should be a good use of one's CPU's idle clocks:
https://en.m.wikipedia.org/wiki/Tijd...njecture#Prize

technically all even powers can be expressed as a square so all even powers can be reduced to x,y or z, being 2 which isn't allowed so you can reduce to odd powers and then since each odd power that isn't prime can be made up of prime powerings you can restate it to be related to prime powers potentially. that's a good starting point I think. oh and of course any x,y,z with common factors are out as then you can restate it as fermat's last theorem as talked about in the page as well.

[a1call]

Well it pays more than the 1 billion digits prime prize.
My 2 cents:
If the addends are coprime, then the sum can only factor into primes which are not any of the prime factors of any of the addends. Infinite examples exist of the sum being multiples of independent from the addends primes. All you got to do is to find a power form. Seems possible except that so does the OP which is proven impossible.

[a1call]

I can see a vagueness with the conjecture. The constraints are not complete as sentenced:

If
A^x +B^y = C^z,
where A, B, C, x, y, and z are positive integers with x, y, z > 2, then A, B, and C have a common prime factor.

If a solution is found with (x || y) == 1 & z>2
Then the constraints can be satisfied by raising the 2 sides of the equation to any integer power greater than 2.

Would x=1, y=1, z>2 results be accepted or rejected?

ETA: I see my mistake its a sum and not a product.
Please disregard this post.

[science_man_88]

Originally Posted by a1call

I can see a vagueness with the conjecture. The constraints are not complete as sentenced:



If a solution is found with (x || y) == 1 & z>2
Then the constraints can be satisfied by raising the 2 sides of the equation to any integer power greater than 2.

Would x=1, y=1, z>2 results be accepted or rejected?

they would likely be rejected since they don't fit x,y,z >2

the cases I was talking about being eliminated already by the constraints and a few other things are:

1) A^(2x) = (A^x)^2 so that eliminates all even exponents
2) if A^(ax)+B^(bx)=C^(cx) then by 1 above (A^a)^x+(B^b)^x = (C^c)^x which is a form equivalent to Fermat's last theorem with x>2. this eliminates all groups where all three exponents share a common factor.



(a+b)^2 = a^2+2ab+b^2 when does this become power ?