<s>[INST] Obfuscated proofs I am, just for fun, looking for long and complicated proofs for statements which can be proven rather easily and much faster. The proof itself still has to be correct however.
While the proof should be obfuscated, all parts should have some relevance. So do not prove Fermat‘s last theorem and end with "ah, by the way: 1+1=2, so the statement follows.
It is also boring to obfuscate simple arithmetic; one can prove "1+1=2" in 100 pages only using addition, subtraction, multiplication and division – but that is not fun.
I rather look for some very interesting obfuscation of a proof. Maybe a statement of elementary number theory can be proven in a "nice" complicated way. Or maybe one can use functional analysis to prove basic analysis stuff etc.

Here is one (not that good) example: Theorem: For $a, b \in \mathbb{R}$ it holds that $(a-b)^2 = a^2 - 2ab + b^2$.
Proof: Let $f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto x^2$. As $f$ is analytical the Taylor expansion of $f$ converges. Therefore
$$
\begin{align*}
    x^2
 &= f(x) \\
 &= Tf(x,b) \\
 &= \sum_{n=0}^\infty \frac{f^{(n)}(b)}{n!} (x-b)^n \\
 &= \frac{b^2}{0!} + \frac{2b}{1!} (x-b) + \frac{2}{2!} (x-b)^2 + \sum_{n=3}^\infty \frac{0}{n!} (x-b)^n\\
 &= b^2 + 2bx - 2b^2 + (x-b)^2 \\
 &= -b^2 + 2bx + (x-b)^2,
\end{align*}
$$
ie.
$$
    x^2 + b^2 - 2bx = (x-b)^2
$$
and for $x = a$ the theorem follows.
 [/INST] Proof: $0a = 0$
$$(a+0)^2 = a^2 = a^2 + 2(0) + 0^2$$
$$a^2 = a^2 + 2(0) + 0^2$$ $$0 = 2(0) + 0^2$$ $$2(0) = (1+1)0 = 1(0) + 1(0) = 0 + 0 = 0$$ $$\dfrac{0^2}{2} >= \sqrt{0^2} = 0$$ $$0^2 >= 2(0)$$
We have proven that $$2(0) = 0$$
Therefore, $$(a+0)^2 = a^2 = a^2 + 2(0) + 0^2 = a^2 + 0$$ $$a^2 = a^2 + 0$$ $$0 = 0$$
Q.E.D.
</s>