Modulo of modulo. Let $p,N$ be positive integers with $N$ divides $p$. Does for every integer $X$, $[X\pmod{N}]\pmod{p}=X\pmod{p}$?

Question

Let $p,N$ be positive integers with $N$ divides $p$. Does for every integer $X$, $[X\pmod{N}]\pmod{p}=X\pmod{p}$?

This question is similar to [1] - I consider it different due to $N$ now dividing $p$ instead.

I am trying to simplify the expression: $$\phi(n)=[(4n)*((4n)^{-1}\pmod{65537})]\pmod{65537*4*n}$$ where $\phi(n)$ is Euler's totient function, and I'm thinking of using the result above for simplifying the RHS.

[The particular example using 65537 I have derived using congruences on the RSA system and application of the CRT.]

Please state what the correct result is and why it's correct. [1][Modulo arithmetic (modulo of modulo)

No, but we have $[X\pmod{N}]\pmod{p}=X\pmod{N}$ (holds in general when $N\leq p$) — Jakobian, Apr 28 '19 at 11:06
Using modulo, we get a number less than that, so when $N\le p$, $X\pmod{N}<N\le p$. — tarit goswami, Apr 28 '19 at 11:09
$X\pmod{N}$ is the rest from division of $X$ by $N$. Then $0\leq X\pmod{N}<N\leq p$ by definition. But then it's the same as the rest from division of $X\pmod{N}$ by $p$ by definition i. e. $[X\pmod{N}]\pmod{p}$. — Jakobian, Apr 28 '19 at 11:14
Ok, thus the correct answer is $[X\pmod{N}]\pmod{p}=X\pmod{N}$ ? — unseen_rider, Apr 28 '19 at 11:23

Bill Dubuque · Answer 1 · 2019-04-28T13:16:52.047

0

If $\,\gcd(4n,p) = 1\,$ then by CRT = Chinese Remainder Theorem (or Easy CRT) we have

$$\begin{align} &x\equiv 0\pmod{4n}\\ &x\equiv 1\pmod{p}\end{align}\,\iff\, x\equiv 4n((4n)^{-1}\bmod p) \pmod{\!4np}$$

Or, expressed in operator form, using the mod Distributive Law

$$ x\bmod 4np \,=\, 4n((4n)^{-1}\bmod p)$$

Generally there is no way to further simplify that other than computing the inverse.

The titled equation is not generally true, i.e. generally

$$ (x\bmod N)\bmod NK \,\neq\, x\bmod NK$$

For example when $\,x = N\,$ it becomes $\, 0 \neq N\bmod NK,\,$ true for all $K>1$ (and $N\neq 0)$

edited Apr 28 '19 at 13:16

answered Apr 28 '19 at 13:11

Bill Dubuque

272,048

how about using $a^{p-2} \cong a^{-1} \pmod{p}$, so $(4n)^{p-2} \cong (4n)^{-1}$ - does this simplify the result? – unseen_rider Apr 29 '19 at 10:48
@unseen_rider That combined with repeated squaring is one way to compute modular inverses, but usually it is more efficient to use the Extended Euclidean Algorithm. – Bill Dubuque Apr 29 '19 at 19:28
Ok, so does this mean that the above $4n[(4n)^{−1}(modp)]$ simplifies to $4n[(4n)^{p-2}(modp)]=(4n)^{p-1}(modp)$ ? – unseen_rider Apr 30 '19 at 20:17
@unseen_rider No, $, x\equiv (4n)^{\large p-1}!\pmod{!4np},$ but that's not "simpler" - it has larger modulus $,4np > p.,$ The point of using $, 4n((4n)^{-1}\bmod p)$ is that we have reduced the modulus from $,4np,$ to $,p,,$ which simplifies the computation (e.g. consider $,p,$ very small compared to $n).,$ Ditto for $,4n((4n)^{\large p-2}!\bmod p),$ but, as I said, generally it is more efficient to compute the inverse via extended Euclidean vs. raising to power $,p-2.,$ – Bill Dubuque May 01 '19 at 01:48

Modulo of modulo. Let $p,N$ be positive integers with $N$ divides $p$. Does for every integer $X$, $[X\pmod{N}]\pmod{p}=X\pmod{p}$?

1 Answers1