I am trying something with Fermat's last theorem:
Maybe I am totally wrong about this and so I wanted to post it here for you guys to try and check it.
Fermat's last theorem states that you will not find any $x,y,z \in \mathbb{N}$ that satisfy : $x^n + y^n = z^n$ where $n>2$ and $n \in N$ as well.
Here is what I thought about:
let $n=t+k$
substitute this to the equation gives us:
$x^{t+k} + y^{t+k} = z^n$
$x^t x^k + y^t y^k = z^n$
$x^t x^k + y^t y^k = x^n + y^n$
$x^n ( x^{t+k-n} -1) = y^n(1-y^{t+k-n})$
now, I know that $t+k-n$ is always $0$
and so no matter which number we choose we get that $1 + 1 = z^a$
$2=z^a$ has no solutions for $z,a \in \mathbb{N}$
so we get that no numbers satisfy fermat's last theorem.
Even if there were x,y that satisfy this, $1-x^{m}$ would be negative while $y^{m} -1$ would be positive.
But I can't see the mistake here…
Best Answer
Since $t+k-n=0,$ $x^n(x^{t+k-n}-1)=y^n(1-y^{t+k-n})$ means $x^n\times 0=y^n\times 0.$ I think you want to say $x^n=y^n$ from here, but that of course is not correct.