understand the difference.?

Question

Sir fermats last theorem states that there is no integer solution for x,
y and z in the equation x^(n)=z^(n)-y^(n) for n greater then 2

My statement is this that there are infinitely many solutions for z and y for a given x and n when x^(n)=z^(2)-y^(2).Please do not mix it with FLT or pythagoras theorem.The mathematical proof of this statement was very helpful for me in solving FLT on a single page by the formula of z^(n)-y^(n) along with the fact that there lies a common term between x,y and z when x+y=z that is n=1 in Sir fermats last theorem (not greater then 2)

i did not said non integer solutions are not invited for y and z but it should be terminating decimals.

Answer 1

Your claim that there are infinitely many solutions for x and y for a given x and n does not hold if we consider only integer solutions.

Consider that we have:
x^n = z^2-y^2
x^n = (z-y)(z+y)
Since z, y are integers, |z-y| >= 1 and |z+y| >= 1
Also, |z-y| <= x^n and |z+y| <= x^n
We can assume that z,y > 0, since if there is a solution with one of them less than 0, then switching the sign will still be a solution.
So, now we know that |z+y| <= x^n. For z,y >=0, this has x^n + 1 solutions (not all of these will solve the original equation, but there are no others that do).

Thus, there are at most 4x^n +4 unique solutions to this equation in the integers. The others are found by taking a positive pair (z,y) and noticing that (-z,y) (z,-y), (-y,-z) are also solutions.

However, if you are talking about non-integer solutions, then there are obviously infinitely many, and that result is not useful.

Answer 2

I'm not sure I understand what you are saying, but assuming you are talking about positive integers, I don't think that there would be an infinite number of solutions for all given x in the above statement. If x=1 there are no solutions for any n. If x=2 and n=3, the only solution I see is z=3 and y=1.

Answer 3

11?