Simple but counterintuitive math

The Basel Problem. Why is the following equation true: $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ I know how to prove this using Fourier Series, but the proof doesn't "preserve intuition" very well.

Selmer's cubic curve. Selmer has proved that the following cubic has a solution modulo $m$ for every positive integer $m$, but does not have any integer solutions: $$3x^3 + 4y^3 + 5z^3 = 0$$ That makes this a counterexample to a property satisfied by homogenous conics, stating that they have a solution in integers if and only if they have a solution modulo each positive integer $m$. Is there a quick or intuitive way to understand why this fails for this cubic?

The quantum CHSH game. This is a very simple cooperative game for two players, Alice and Bob, who are allowed to discuss strategy before playing the game separately from each other. The game has a moderator named Charlie, and the game proceeds as follows:

1. Charlie chooses $x,y\in {0,1}$ uniformly and at random.
2. Charlie sends $x$ to Alice and $y$ to Bob.
3. Alice must choose a bit $a$, and Bob must choose a bit $b$.
4. Alice and Bob win the game iff $a + b\bmod 2 = x\land y$.

Classically, it is easy to prove that the optimal strategy allows them to win $75\%$ of the time. But when quantum strategies are permitted, so that Alice and Bob may share an entangled quantum state, the following winning probability is realizable, and furthermore optimal: $$\cos^2\tfrac{\pi}{8} \approx 0.85$$ This is unintuitive to me in two ways: firstly, that such a strategy is possible in the first place; and secondly, that it cannot be improved on by adding "more entanglement" somehow. I've proven both of these things algebraically, but they still elude my intuition (as many quantum-related things do).

Exotic S5-shaped subgroups of S6. There are 6 conjugate subgroups of $S_6$ that are isomorphic to $S_5$, and that act transitively upon a 6-element set as subgroups of the permutation group representation of $S_6$. (That is, they are not the trivially embedded copies of $S_5$ that simply fix one of the 6 elements.) Is there an intuitive - perhaps geometric or algebraic - way of understanding these subgroups?

Magma theories over finite sets. Terence Tao's "Equational Theories" formalization project holds many gems of counterintuitive mathematics, but one of the things that intrigues me most is the idea that some equational laws, called Austin laws, admit only infinite (or trivial) models. Examples include:

$$\begin{align*}x &= y\circ (y\circ (y\circ (x\circ (z\circ y)))) \\ x &= (y\circ ((z\circ x) \circ w))\circ (x\circ w) \\ x &= (((y\circ y)\circ y)\circ x)\circ ((y\circ y)\circ z)\end{align*}$$

It is also interesting how some pairs of magma laws have an entailment relationship for finite magmas, but not for infinite magmas. I find both things very counterintuitive but haven't had the time to explore the topic enough for them to become more clear to me.

Equational proofs can be ridiculously hard. There are many interesting problems that take the following form: given that some equations hold true in a ring/magma/monoid/group/field, show that some other equation holds true in general, or that some other equation has a solution. One of the most appealing kinds of proofs for these problems is that consisting of direct algebraic manipulation. For example, a favorite problem is to prove that $$X = 1 + X^2 ~ \implies ~ X = X^7$$ in, for instance $\mathbb C$. One proof is to show that $X$ is a cube root of unity, from which the result is straightforward. But there is a more interesting and more generalizable proof that does not even rely on the field properties of $\mathbb C$, and in fact applies in more general settings (one consequence, for example, is that a tree-like topological space satisfying $T\simeq \mathbf{1} + T^2$ also satisfies $T \simeq T^7$).

But sometimes, direct algebraic proofs are shockingly way more complicated than constructive ones. For example, Jacobson's Theorem states that rings $R$ for which $x^n = x$ for all $x\in R$ for some $n\in\mathbb N$ are commutative. Direct algebraic evidence for this claim for a particular value of $n$ can be provided by expressing the commutator $xy-yx$ in the following form: $$xy-yx = \sum g_i\cdot (f_i^n - f_i)\cdot h_i$$ for some $f_i,g_i,h_i\in \mathbb Z[x,y]$. But finding these representations is difficult and they can be ridiculously complicated for even very small values of $n$. See this paper for details.