Let $F = (\Bbb{R}, \oplus_d, \cdot)$ be the field with usual $\cdot$, and $\oplus_d$ is defined as $a \oplus b = (\sqrt[d]{a} + \sqrt[d]{b})^d$. This field is isomorphic to usual $\Bbb{R}$ structure by $f: \Bbb{R} \to F, \ x \mapsto x^d$, $d$ an odd prime.
Since there are no positive integer solutions to $x^n + y^n - z^n = 0$, for $n \gt 2$, then for $d$ odd prime we have that $(x,y,z)$ is a solution if and only if $f(x^n + y^n -z^n) = 0 = x^{dn} \oplus y^{dn} \oplus -z^{dn}$, i.e. iff it's also a solution to that equation in $F$. Then in summary, for all $n \gt 2$ and odd primes $d$, $x^{dn} \oplus y^{dn} \oplus - z^{dn} = 0$ has no solution.
Further, we have that $g: F \to F \ , x \mapsto x^d$, is a ring homomorphism. Then we have that there are no solutions to $x^{d^k n} \oplus y^{d^k n} \oplus -z^{d^k n} = 0$ for all $k \geq 1$.
I just want to verify if this is all true and if you notice any interesting properties, please mention. Thanks.
Consider the Beal equation $x^a + y^b - z^c = 0$. Let $a,b,c$ be odd primes. $(x,y,z)$ is a solution to it iff $(x,y,z)$ is a solution to $x^{d^k a} \oplus y^{d^k b} \oplus -z^{d^k c} = 0. \ $ So then using the inverse isomorphism $x \mapsto \sqrt[d]{x}, \ F \to \Bbb{R}$. We have that $(x,y,z)$ is a solution to the Beal equation iff it's a solution to $(x^{d^{k-1}a} + y^{d^{k-1}b} - z^{d^{k-1}c})^d = 0$ and since we're in an integral domain we can take the inside of that exponential equal to $0$.
Is that true? Thank you!
Edit So we have that $(x,y,z)$ is a solution to the Beal equation iff it's a solution to
$$x^{d^k a} + y^{d^k b} - z^{d^k c} = 0$$
for $k \geq 0$. But that's infinitely many solutions to FLT which is impossible.
That proves the Beal conjecture, but probably not! Where is my error?
According to Bruno Joyal in an answer down below. $x \mapsto x^d$ is not an automorphism of $F$. Indeed it is:
Clearly since it's the $d$th root function for odd $d$ it's a multiplicative bijection of the set $\Bbb{R}$ onto itself. That's easy enough to show though: It has an inverse given by $x \mapsto \sqrt[d]{x}$. So it's a bijection. It's a homomorphism too: clearly it's multiplicative. So let $x, y \in F$, then $(x \oplus y)^d = (\sqrt[d]{x} + \sqrt[d]{y})^{d^2} =(\sqrt[d^2]{x^d} + \sqrt[d^2]{y^d})^{d^2} = x^d \oplus_{d^2} y^d $. So it's an isomorphism into $(\Bbb{R}, \oplus_{d^2}, \cdot)$. Didn't see that. :D
What you have done is merely relabel the elements of $\mathbf R$ by using the fact that $x \mapsto x^d$ is a bijection of $\mathbf R$ with itself when $d$ is odd. This cannot lead to new information (such as new solutions to equations). Your mistakes begin when you say "use the inverse isomorphism"; you should be getting the same old equation that you started with back, not the apparently new equation.