A to the n plus b to the n cannot equal c to the n plus d. Are there any d's besides zero that we know this to be true for any exponents?

So i suppose you are asking what is known about the generalization of FLT in the form;

x^n+y^n=z^(n+d) with d not n*k x,y,z>0


At least x^3+y^3=z^2 has solutions;

1^3+2^3=3^2 is one.

11^3+37^3=228^2 is another.


Here is a lecture on some generalized versions of Fermat's equation.

http://math.hawaii.edu/numbertheory2...tt-Lecture.pdf

A good starting point is probably to always check the corresponding elliptic curves that were used to make the original Fermat problem connection eg by Frey and Hellegouarch

like here https://en.wikipedia.org/wiki/Frey_curve

you may check out info found here

http://arxiv.org/pdf/math/9905208.pdf

and also here

https://en.wikipedia.org/wiki/Fermat...lan_conjecture

Regarding that last one as of 2015 only known;
As of 2015, the following ten solutions to (1) are known:[1]

1^m+2^3=3^2
2^5+7^2=3^4
13^2+7^3=2^9
2^7+17^3=71^2
3^5+11^4=122^2
33^8+1549034^2=15613^3
1414^3+2213459^2=65^7
9262^3+15312283^2=113^7
17^7+76271^3=21063928^2
43^8+96222^3=30042907^2

No sorry. d was added to the total, not the exponent. Eg a fifth power plus a fifth power minus d isn't a fifth power.

So you want to know if the equation

x^n+y^n-z^n=d has no solutions for certain d (like we know it doesnt have for d=0)

eg if d=1 it does have the x,1,x trivial triplet solution.

But if d=2 you ask if for example there is any x,y,z and n>1 (or n>2 if its trivial at n=2 havent checked yet) that satisfy;

x^n+y^n-z^n=2


Eg the d=3, n=3 case is eliminated since

4^3+4^3=5^3+3

Also the n=2, d=3 is eliminated as

4^2+6^2=7^2+3.

Before finding a d for all exponents how about for just one exponent. Is there any exponent where we know that the sum of two of those powers never differs from another of those powers by d?

It looks like d=2, like the original Fermat equation, is impossible to solve for n>2. (not a proof yet if ever, a likely conjecture)

It also looks like x^2+y^2=z^2+d has infinite number of solutions for any d.

eg if d=2, n=2 we can choose any z= ((2k+1)^2-1)/2, x=(2k+1)^2-3)/2, y=2k+1 for any natural number k (and there are more than that kind of solution actually)

Also it looks like any d>1 will in general not lead to solutions if d is small enough you know below the obvious x,y,x+1 choices y<x when d happens to be difference of these obvious trial choices.

eg if we have selected z=x+1 and consider (x+1)^n-x^n= (x^(n-1)+...+1) and then take y= integerpart((x^(n-1)+...+1)^(1/n)+1)

we can see obviously that when -d = (x+1)^n-x^n-integerpart(((x+1)^n-x^n)^(1/n)+1)^n we have a solution for that d

example

if n=3 and say x=5 then 5^3-4^3=61=64-3=4^3-3 or 4^3+4^3=5^3+3 giving us one solution for d as small as 3

if n=4 and say x=7 we have 7^4-6^4=1105=6^4-191 so 6^4+6^4=7^4+191 so for d=191 we have a solution etc.

Only integers allowed?

Perhaps explore parallel ideas in Ramanujan's tau function.

There might be if either b or c is the centre of a circle. Perhaps a digression for this particular theorem but I think there is some ore to be exposed in seeing what new questions will yield out of this and the other conjectures.

Probably not a static solution. Which makes it unacceptable. It could be a decent drop and add algorithm however.

i.e. if a to the n, then b to the n (for n defined by how b is to a in relation to c, which will suggest a range for d. if a zero is self-evident, then n might be both positive and negative dynamically.)

Uh, or not.

I havent checked and it might be fun to search this if the generalized powers conjecture is positive though;

By that i mean lets try to see if there is always a non trivial solution (not all but 2 zeros i mean or usage of negative integers) to the problem;

x1^n+x2^n+..xn^n=y^n


Funny how 3^3+4^3+5^3=6^3 (thats amazing actually lol)

So does for example the sum of 4 4th powers equal a 4th power eventually anywhere? Then the sum of 5 5th powers etc.

For n=4 we have

95800^4 + 217519^4 + 414560^4 = 422481^4 which uses one as 0 which is ok with me (or one may want to insist on all n non zero but i dont think that is important)

If it fails which is the first n that it fails and if it can be proven of course.