# Fermat's last theorem. Where is the mistake?

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...

• You lost me after "$t+k-n$ is always $0$". Could you elaborate what you have done? – Ishan Deo Jun 9 at 19:17
• @IshanDeo No matter which numbers we choose we get that 1-1=0. while that is true, I don't see how we can pick x,y and and get that $x^0 + y^0 = z^0$ (or $z^a$).. that just breaks the rule that $z \in \mathbb{N}$ or that $a \in N > 2$ – dexamol Jun 9 at 19:20
• Since $t+k-n=0$, we have $x^{t+k-n}=y^{t+k-n}=1$ so your equation just reads $0=0$. Not sure what you hope to conclude from that. – lulu Jun 9 at 19:23

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.