Explaining modulo reduction of curve25519, multiply hi bits with 38 trick? I have learnt that there is a trick where you can speed up the reduction modulo of a point (x-value) in a x25519 curve. Since, it uses the prime number $2^{255} - 19$.

From article: 


  Reduction modulo exploits the fact that $2^{255} - 19$. $2^{256} \equiv 38$, so $38*r4$ is added $r0$ and $38*r5$ is added $r1$ and so on. 


$r_n$ are registers with size 64 bits.

The problem I have. I don't really understand why this is the case? I wonder if anyone can expand on this?
Let $p = 2^{255} - 19$.

Clearly $p \equiv 0 \pmod p$, meaning $p$ (the modulus) divides $p - 0$ (the two sides of the equation), or equivalently: there exists some integer $k$ such that $p - 0 = k\cdot p$.  (Here $k = 1$.)

So $2^{255} - 19 \equiv 0 \pmod p$, and thus $2^{255} \equiv 19 \pmod p$, meaning there exists some $k$ such that $2^{255} - 19 = k\cdot p$.  (Here, again, $k = 1$.)

If we multiply both sides of the equation by $2$, then we get $2\cdot 2^{255} \equiv 2\cdot 19 \pmod p$, meaning there exists some integer $k$ such that $2\cdot 2^{255} - 2\cdot 19 = k\cdot p$.  (Here $k = 2$.)

But this equation is just $2^{256} \equiv 38 \pmod p$.

Hence, whenever you are doing arithmetic modulo $p$, the quantities $2^{256}$ and $38$ are equivalent.  Since reduction modulo $p$ is a ring homomorphism—that is, $$[(a + b) \bmod p] \equiv [(a \bmod p) + (b \bmod p)] \pmod p,$$ and likewise with multiplication—we can split a number $n$ into the low 256 bits $n_{\mathrm{lo}}$ and the rest $n_{\mathrm{hi}}$ so that $n = n_{\mathrm{lo}} + 2^{256} n_{\mathrm{hi}}$.  Then $$n = n_{\mathrm{lo}} + 2^{256} n_{\mathrm{hi}} \equiv n_{\mathrm{lo}} + 38 n_{\mathrm{hi}} \pmod p.$$