While I was surfing the web, searching things about math, I read something about a particular theorem of Fermat. It said: let $a$ and $b$ be rational. Then $a^4-b^4$ cannot equal the square of a rational number, so $a^4-b^4\neq c^2$ with $c$ rational. My question is: did I understand the theorem? if not, what is the actual theorem? if yes, how can I proof this or is that too difficult?
Yes, that is a theorem, No, it’s not too difficult to prove, and you can try in some analytic number theory book for the proof, say in
That’s right, modulo a small precision: Fermat proved that there are no solutions other than the obvious ones, such as $(a,b,c)=(0,0,0)$, $(1,0,1)$, etc.. See for instance this wikipedia page.
Yes you understand the theorem. The proof uses the method of “infinite descent” (a sibling of reductio ad absurdum, or proof by contradiction).
There is a walkthrough proof here:- https://en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#Application_to_right_triangles
But did you maybe want just a hint instead??
If $b\neq 0$, yes, it’s true. It follows from the fact that
$$ a^4-b^4=c^2 $$
has no integer solutions, under the assumptions $\gcd(a,b)=1$ and $b\neq 0$.
On the other hand, if $c,d$ are two coprime integers such that $c^2+d^2$ is a square, say $e^2$, than $e$ is the sum of two coprime squares. This is the key step of the proof.