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Given prime $p, 0 < m < p$ and $ed \equiv 1 \pmod{p-1}$, prove $ m^{ed} \equiv m \pmod{p}$.

I get that this is hinting at a proof very similar to that of RSA, and that I have to consider when $\gcd(m,p)=1$ and when it doesn't. I also know that I need to use Euler's theorem and CRT. I just can't get past the $p-1$ passing itself into the mod from Euler's theorem. How should this proof look?

Arturo Magidin
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Christheyankee
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  • Your question will be better recieved if you write down your proof or at least parts of it and ask specific questions about it. – Vinyl_cape_jawa Feb 23 '19 at 23:00
  • Euler’s Theorem tells you that $x^{\varphi(n)}\equiv 1\pmod{n}$ when $\gcd(x,n)=1$. If $\gcd(x,n)=1$, and $r\equiv 1\pmod{\varphi(n)}$, then $r=1+k\varphi{n}$ for some integer $k$, so $x^{r} = x^{1+k\varphi(n)} = x^1x^{k\varphi(n)} = x(x^{\varphi(n)})^k$. What does this equal modulo $n$? And what is $\varphi(n)$ when $n$ is a prime? – Arturo Magidin Feb 23 '19 at 23:05
  • First question : m mod n as the proof requests. But don't I have r = 1 mod (p-1)? And idk when n is prime. @ArturoMagidin – Christheyankee Feb 23 '19 at 23:11
  • I do not understand anything of what you wrote. You say “first question”, but what follows is not a question, it’s a sentence fragment. Don’t confuse the letters I wrote with the letters in your original problem. I’m using $n$ and $x$ as arbitrary letters, not related to your problem statement. Did you understand what I wrote? And do you really don’t know the value of Euler’s phi function when the input is a prime? If that is the case, then sorry, but you are too far behind to tackle this problem. You need to go way back and actually learn the stuff first. – Arturo Magidin Feb 23 '19 at 23:13
  • @ArturoMagidin okay I meant the phi function of p when p is prime is p-1. But idk where to go with that. You did the proof for when gcd(x,n) = 1 but Idk what to do when it isn't/ – Christheyankee Feb 23 '19 at 23:18
  • Look at your problem. $p$ is a prime. $m$ is a positive integer smaller than the prime. What is the gcd of $m$ with $p$? – Arturo Magidin Feb 23 '19 at 23:19
  • oh lol it's always one right? – Christheyankee Feb 23 '19 at 23:23
  • Are you asking me or telling me? – Arturo Magidin Feb 23 '19 at 23:24
  • Telling because factors of a prime are one and itself and if the number is less than the prime in order for the gcd not to be one the number would have to be a factor of the prime which it cant be by the definition of a prime. – Christheyankee Feb 23 '19 at 23:26
  • So, presumably, you’ve got your hints and hnnow how to solve the problem. Great. May I suggest posting your final solution as an answer? That way the question won’t go “unanswered”, and other people can help and point out any gaps or errors. – Arturo Magidin Feb 23 '19 at 23:27

2 Answers2

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First consider gcd(m,p). p is prime and p > m.

Conclude gcd(m,p) = 1.

By Euler's Theorem $ m^{\phi(p)} \equiv 1 \mod p, ed \equiv 1 \mod \phi(p), ed \equiv 1 +k\phi(p),k$ is a constant integer and then $(m^{e})^d = m^{ed} \equiv m^{1+k\phi(p)}\\ \\ m^{ed} \equiv m *m^{k\phi(p)}\\m^{ed} \equiv m *1\mod p\\m^{ed} \equiv m \mod p$

Christheyankee
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  • Use \pmod to get the proper notation and spacing, not \mod.Look at the edits I did to your question to see the sytax and formatting. – Arturo Magidin Feb 23 '19 at 23:49
  • Your first paragraph is wrong. You do need to consider it; it’s just that it is equal to 1. And therefore, you do not suppose that $\gcd(m,p)=1$. You conclude that it is one because $p$ is prime and $m$ is positive and strictly smaller. You never say what $k$ is (not any constant will make this work, it has to be a specific constant). Your second line is wrong (where did that addition symbol come from?) – Arturo Magidin Feb 23 '19 at 23:51
  • Thanks, better? @ArturoMagidin – Christheyankee Feb 23 '19 at 23:55
  • $P$ and $p$ will be understood to mean different things, so you shouldn’t use them interchangeably. (If that was your interface automatically changing it to an upper case because if tollows a period, ignore that comment). You still never say what $k$ is or why it matters, and you still have a plus sign that doesn’t make any sense in your second equation. – Arturo Magidin Feb 24 '19 at 00:00
  • Yeah p was automatically capitalized sorry about that. Fixed the plus sign, not sure what k is missing. – Christheyankee Feb 24 '19 at 00:07
  • In line 3 you say, “$k$ is a constant integer”. But you never defined $k$ as something specific, so right now it’s just an arbitrary integer. But if it is an arbitrary integer, then why do you claim that $m^{ed}\equiv m^{1+k\phi(p)}$? Since you did not specify the integer $k$, you are asserting that this holds for any positive integer $k$. So if I plug in $k=3$, it’s true; if I plug in $k=73$ it’s true. If I plug in $k=4128973126321236$, it’s true. Or at least, that’s what you are claiming happens. Does it? (And, congruent modulo what?) – Arturo Magidin Feb 24 '19 at 00:11
  • I think they want you to say that the existence of a k is implied by the previous modulus (and moreover that it doesn't work for just any k). – Boots Feb 24 '19 at 00:13
  • I know you have to work mod phi (p) in the exponent but beyond that I'm not sure. K >= 1? @ArturoMagidin – Christheyankee Feb 24 '19 at 00:23
  • @Christheyankee: Re-read my original comment and try to follow the substance; don’t just try to emulate the form. In particular, there was a $k$ I used, but it wasn’t just any $k$; it was a $k$ whose existence and properties were implied by another assumption I had. – Arturo Magidin Feb 24 '19 at 00:43
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    Did I add the assumption correctly? @ArturoMagidin – Christheyankee Feb 24 '19 at 00:53
  • You might want to say that $ ed \equiv 1 \pmod{\phi(p)}$ implies the existence of an integer k such that $ed = 1 + k \phi(n)$ – Boots Feb 24 '19 at 01:09
  • @Christheyankee: You did not add any assumption, so.. no. Look at Boots comment. Look at my original comment. Where did $k$ come from, and what properties did it have? They are key. You are just ignoring them. – Arturo Magidin Feb 24 '19 at 01:19
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It's a special case of this mod exponent reduction law (crucial to master for problems like this)

Lemma $\ \bmod n\!:\,\ \color{#c00}{a^{\large k}\equiv 1}\,\Rightarrow\,a^{\large j}\equiv a^{\large j\bmod\color{#c00} k}$

Proof $\ $ Dividing $\ j\div k\,\Rightarrow\, j = r + k\,q\ $ for $\ r = j\bmod k = $ remainder, and $\,q = $ quotient,

hence $\bmod n\!:\,\ a^{\large j}\equiv a^{\large r+kq}\equiv a^{\large r}(\color{#c00}{a^{\large k}})^{\large q}\equiv a^{\large r}\color{#c00}1^{\large q}\equiv a^{\large r}$

Thus $\bmod p\!:\,\ \color{#c00}{m^{\large p-1}\equiv 1}\,\Rightarrow\, m^{\large ed}\equiv m^{\large ed\bmod\color{#c00}{p-1}}\equiv m^{\large 1}$

Bill Dubuque
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