I'm following this (Chinese Remainder Theorem and RSA) post, but I don't understand how

$(c^d \bmod n) \bmod p = c^d \bmod p$.

Being told that it's because $n=pq$ is not enough for me to understand.

Could someone please elaborate on that to help me fully understand what's going on here?

Any input would be GREATLY appreciated!


1 Answer 1


Well, lets see if we can go through it from the basics.

$x \bmod y$ is the unique integer $x + \ell y$ that satisfies $0 \le x + \ell y < y$, for some integer $\ell$ (which might be positive, negative or zero) I will skip the parts that prove that, if $x, y$ are both integers and $y > 0$, then there exists such an $\ell$ and it is, in fact, unique.

Now, when we have $(x \bmod pq) \bmod p$, we can replace the inner $(x \bmod pq)$ with $x + \ell_1 pq$, we get $x + \ell_1 pq \bmod p$. This in turn becomes $x + \ell_1 pq + \ell_2 p$. We further note that $\ell_4 = \ell_1 q + \ell_2$ is an integer (which we will denote at $\ell_4$) [1], and so we have:

$$(x \bmod pq) \bmod p = x + \ell_4 p$$

On the other side, $x \bmod p$ is the same as $x + \ell_3 p$, that is,

$$x \bmod p = x + \ell_3 p$$

So, on one side, we have $x + \ell_4 p$, which is a value between 0 and $p-1$.

On the other side, we have $x + \ell_3 p$, which is also a value between 0 and $p-1$.

Since both $\ell_3$ and $\ell_4$ are integers, these must be the same, and hence the original values $(x \bmod pq) \bmod p$ and $x \bmod p$ must also be the same.

Then, replace $x$ with $c^d$ and $pq$ with $n$, and there you go...

[1]: BTW: this is the step where we assume that $p$ is a factor of $n$ if that is not true, then relation will not hold in general.

  • $\begingroup$ Thanks a lot for your help! :D $\endgroup$ Commented Jul 9, 2020 at 19:09

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