So i am wondering how i can figure out what the functional inverse of $x^3$ mod $55$ is.
I can only assume it is $x^{1/3}$ mod $55$ but i am not sure if that is the form i should keep it in
Asked
Active
Viewed 511 times
3
-
1Your question is ambiguous. Do you want the inverse function or the reciprocal (the multiplicative inverse)? – Robert Israel May 06 '16 at 00:44
-
Do mean multiplicative inverse (as you have written) or functional inverse (as in cube root)? – lulu May 06 '16 at 00:44
-
Not all cubes have an inverse mod $55$. – Bernard May 06 '16 at 00:46
-
In either case, Euler's theorem will be useful. – Robert Israel May 06 '16 at 00:46
-
I am looking for the inverse function, i guess i mixed things up in my brain here. – user2327195 May 06 '16 at 00:47
-
I am curious how Euler's theorem factors into the inverse function here. – user2327195 May 06 '16 at 00:54
-
@user2327195 You should make sure that you understand the explanation in the first part of my answer since that is the key idea in many problems like this. I emphasized that since it is not explained at all in the other answer (where, alas, the inverse is pulled out of a hat, like magic). – Bill Dubuque May 06 '16 at 02:59
2 Answers
4
Suppose that $x^3\equiv y\pmod{55}$, where $y$ is relatively prime to $55$. Then $x^{20}\equiv 1\pmod{55}$, since $x^4\equiv 1\pmod{5}$ and $x^{10}\equiv 1\pmod{11}$.
It follows that $x^{21}\equiv x\pmod{55}$. Note that in fact this congruence holds for all $x$. Thus if $y\equiv x^3\pmod{55}$, then $$x\equiv (x^3)^7\equiv y^7\pmod{55}.$$
André Nicolas
- 507,029
-
Im curious, could i use the fact that $\phi(55)$ is $40$ giving $x^{40} \equiv 1$ mod $55$ thus $x^81 \equiv x$ mod $55$ hence $y^{27}$ is also an inverse? I am curious as to how you got $x^{20}$, is this a result from Chinese Remainder Thm or something of the sort? – user2327195 May 06 '16 at 02:22
-
-
1@user2327195 $,\color{#c00}{x^4\equiv 1},\Rightarrow,x^{20}\equiv (\color{#c00}{x^4})^5\equiv \color{#c00}{1}^5\equiv 1\pmod{4},$ for $x$ coprime to $5$ by little Fermat. Similarly $x^{10}\equiv 1,\Rightarrow,x^{20}\equiv 1\pmod{11},$ for $x$ coprime to $11$. So for $x$ coprime to $5,11$ we have $x^{20}-1,$ is divisible by $5$ and $11$ so also by their LCM $= 5\cdot 11,,$ i.e. $,x^{20}\equiv 1\pmod{55}$ You could use CRT insteead of the LCM, but that's overkill. – Bill Dubuque May 06 '16 at 02:45
-
Ah ok. That makes sense. Though i am still curious if $y^{27}$ would also be a correct functional inverse as well? Does this mean functions can have many inverses? – user2327195 May 06 '16 at 02:48
-
1@user2327195 Because $27\equiv 7\pmod{20}$ it follows that $,y^{27}\equiv y^7\pmod{55}$ for all $y$ coprime to $55$, i.e. they are the same functions on the set of such $y$. – Bill Dubuque May 06 '16 at 02:51
-
@user2327195 Btw, you can also use Carmichael's generalization of Euler's theorem to get $,x^{20}\equiv 1$. This gets a sharper result by combining the powers using LCM (vs. multiplying). – Bill Dubuque May 06 '16 at 03:17
2
By the remark below $\,x^{\color{#c00}{\large 20}}\equiv 1\pmod{55}\,$ for all for all $x$ coprime to $55$. Thus all exponents on $x$ can be considered to be $\,$ mod $\,\color{#c00}{20}$.
Solving the equation $\ x^{\large 3}\equiv y\pmod{55}\, $ for $\,x\,$ is easy because $\,1/3\,$ exists mod $20$, namely:
Raise both sides to power $\,\color{#0a0}{1/3}\equiv 21/3 \equiv\color{#0a0}7\pmod{\!\color{#c00}{20}}$
Doing that we obtain $\, x \equiv\, (x^{\large 3})^{\large\color{#0a0}{1/3}}\,\equiv\, y^{\large\color{#0a0}7}\pmod{\color{#c00}{20}} $
Remark $\,\color{#c00}{x^4\equiv 1}\,\Rightarrow\,x^{20}\equiv (\color{#c00}{x^4})^5\equiv \color{#c00}{1}^5\equiv 1\pmod{5}\,$ for $x$ coprime to $5$ by little Fermat. Similarly $x^{10}\equiv 1\,\Rightarrow\,x^{20}\equiv 1\pmod{11}\,$ for $x$ coprime to $11$. So for $x$ coprime to $5,11$ we have $x^{20}-1\,$ is divisible by $5$ and $11$ so also by their LCM $= 5\cdot 11,\,$ i.e. $\,x^{20}\equiv 1\pmod{55}\,$ Alternatively, we could use CRT instead of the LCM, but that's a bit overkill.
Bill Dubuque
- 272,048