All smooth manifolds are triangulable, as you say. This follows from Morse theory, which dictates that you only need to know how to triangulate (PL) handle-attachments, which one can do by hand, and this has been known since the 30s. So in my post I meant "topological manifold" when I said "non-triangulable manifold", because that is the only thing that could have made sense! 

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As for "triangulable iff smooth". The standard statement is that the notion of PL and smooth structures are equivalent in dimensions up to 6 (in dimension 7, every PL structure may be smoothed, but not uniquely). See [here](https://mathoverflow.net/q/96670/40804) for a good discussion. 

A PL manifold can be thought of as a topological manifold $X$ equipped with a(n equivalence class of) triangulation so that the [link](https://en.wikipedia.org/wiki/Triangulation_(topology)#Piecewise_linear_structures) of any vertex is PL-homeomorphic to $S^3$. 

In dimension 4 it is already difficult to prove that "triangulable and PL-able" are equivalent. In any dimension, it is not hard to show that $\text{lk}(\sigma)$ is a [homology manifold](https://en.wikipedia.org/wiki/Homology_manifold) using the cone-homeomorphism from $\text{lk}(\sigma) \times \Bbb R$ to an open subset of a topological manifold; in particular, $\text{lk}(\sigma) \times \Bbb R$ is a topological manifold. It is similarly not hard to show that $\text{lk}(\sigma)$ must have the same homotopy type as $S^{\dim M - 1}$. In dimensions $\leq 2$, homology manifolds are in fact manifolds (Theorem 16.32 in Bredon's sheaf theory) and topological manifolds admit unique PL structures, so we see that there is a PL homeomorphism $\text{lk}(\sigma) \cong S^{\dim M - 1}$. (In fact, because the sphere carries a unique PL structure, one could just replace this requirement with "is homeomorphic to..." instead of "is PL homeomorphic to...") In particular, any triangulation of a manifold of dimension at most $3$ equipped is automatically PL.

In dimension 4 the same is true but one needs to do more work: now you need to use the fact that $\text{lk}(\sigma)$ is itself a simplicial complex. A theorem of Edwards (theorem 3.5 [here](https://www.maths.ed.ac.uk/~v1ranick/papers/cannon2.pdf)) then indicates that $\text{lk}(\sigma)$ is a manifold, and hence by the Poincare conjecture that it is homeomorphic to $S^3$. 

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In dimensions at least 5 you should expect this to be false, because if $P$ is a homology sphere of dimension $n \geq 3$, then $\Sigma^2 P \cong S^{\dim P + 2}$ is naturally triangulated (you may always suspend a triangulation), but it is not a PL structure: the links of the top/bottom vertices are $\Sigma P$, which is not a manifold. Of course, the sphere is PL-able, so this is not a counterexample to your question. But it indicates there's no reason to believe there isn't one - the links of vertices are not automatically spheres for silly reasons anymore.

Now apply [Galewski-Stern](https://www.maths.ed.ac.uk/~v1ranick/papers/galester.pdf), which says the following (though note the discussion of the Kirby-Siebenmann invariant is more classical than their paper!). 

>There is a group $\Theta$, called the homology cobordism group, which fits into a short exact sequence $0 \to \text{ker}(\mu) \to \Theta \to \Bbb Z/2 \to 0$. There is an invariant $\text{ks}(M) \in H^4(M;\Bbb Z/2)$ of topological $n$-manifolds, and if $n \geq 5$ the vanishing $\text{ks}(M) = 0$ is equivalent to $M$ supporting a PL structure. Assuming still that $n \geq 5$ and writing $\beta_{\Theta}: H^4(M;\Bbb Z/2) \to H^5(M;\text{ker}(\mu))$ for the associated Bockstein map, then $\beta_{\Theta} \text{ks}(M)$ vanishes if and only $M$ is triangulable. 

The fact that $\beta_\Theta$ is nonzero is essentially Manolescu's 2012 contribution to the triangulation question.

So what you want to find is an example of a $(\geq 5)$-dimensional $M$ so that $\text{ks}(M) \neq 0$ but $\beta_\Theta \text{ks}(M) = 0$. Take $X$ to be the $E8$ manifold, and let $X_k = X \times T^{4 - k}$ be the $k$-dimensional manifold you get by crossing with enough circles. We already know that $\text{ks}(X) \neq 0$, because for any 4-manifold with even intersection form, $\text{ks}(X) = \sigma(X)/8 \pmod 2$, and $\sigma(X) = 8$ in this case. Because Kirby-Siebenmann invariants of products $X \times Y$ are (under the Kunneth decomposition) the same as $\text{ks}(X) + \text{ks}(Y)$, we see that $\text{ks}(X_k) = \text{ks}(X) \neq 0$ for all $k$. But because $\beta_\Theta H^*(X;\Bbb Z/2) = 0$, and $\beta_\Theta$ acts on $H^*(X \times Y;\Bbb Z/2) = H^*(X;\Bbb Z/2) \otimes H^*(Y;\Bbb Z/2)$ via $\beta_\Theta \otimes 1 + 1 \otimes \beta_\Theta$, we see that $\beta_\Theta \text{ks}(X_k) = \beta_\Theta \text{ks}(X) = 0$ for all $k \geq 4$. 

Thus by the Galewski-Stern result (and Kirby-Siebenmann's work on PL structures), $X_k$ is triangulable for every $k \geq 5$ but not PL.