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...

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    $\begingroup$ You lost me after "$t+k-n$ is always $0$". Could you elaborate what you have done? $\endgroup$ – Ishan Deo Jun 9 at 19:17
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    $\begingroup$ @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 $a \in N > 2$ $\endgroup$ – dexamol Jun 9 at 19:20
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    $\begingroup$ 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. $\endgroup$ – 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.

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    $\begingroup$ Oh, I get it. I'll think of something else. It is really fun to think of some proofs though :) Thanks! $\endgroup$ – dexamol Jun 9 at 19:24
  • $\begingroup$ As long as it is only for fun...But you should be aware that no "elementary proof" (in a precise sense) of FLT can exist. See e.g. math.stackexchange.com/a/1590336/300700 $\endgroup$ – nguyen quang do Jun 11 at 7:03

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