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$$\begin{aligned}m^{ed} &\equiv m\bmod n\\ ed &\equiv 1 \bmod \phi(n)\\ \end{aligned}$$

Why does the modulus of the modular multiplicative inverse have to be the totient function? Won't any positive integer coprime with $e$ work? Can someone explain why the totient function is needed? How did the inventors of RSA arrive at the totient function?

NasuSama
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Leo Jiang
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  • Are you familiar with Euler's Theorem from elementary number theory? – Gerry Myerson Jan 03 '14 at 18:39
  • I know the formula. Won't $m^x \equiv 1 (mod n)$ work for many values of $x$? Not just the Totient function? – Leo Jiang Jan 03 '14 at 18:42
  • Yes, but all those values are (in general) multiples of the totient. – Gerry Myerson Jan 03 '14 at 18:43
  • Do you have any specific example of a reasonably sized $n$ such that $m^x\equiv m\pmod n$ for all (or all coprime) $m$ and where $x$ is not $\equiv 1\pmod {\phi(m)}$? – Hagen von Eitzen Jan 03 '14 at 18:43
  • @GerryMyerson Do you know why? I don't doubt the claim, I just don't understand it. – Leo Jiang Jan 03 '14 at 18:48
  • @HagenvonEitzen No, I'm still trying to understand the basics of modular arithmetics. Logically, there should be. I'll think about it. – Leo Jiang Jan 03 '14 at 18:49
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    @HagenvonEitzen $\lambda(n)+1$, the Carmichael function. – Daniel Fischer Jan 03 '14 at 18:58
  • RSA uses the totient function because it is relatively obvious that RSA works as intended if you use it. Might it work with some other value? Maybe, but why complicate matters? Especially in cryptography, complexity should be avoided whenever possible, it makes algorithms and their security harder to analyse (and there are enough unknowns about RSA as it is). – fkraiem Jan 03 '14 at 19:24

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If ϕ(n) is not used in RSA then the calculation of private key (e) will be very much easy for the attackers (ie) as per RSA ed ≡ 1 mod ϕ(n). Here instead of ϕ(n), if n is used then the attacker can easily calculate e value since d,n are publicly known to all. Hence to assure secrecy in calculating the private key 'e' ϕ(n) is used.

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    Wrong... $\phi(n)$ is essential for the algorithm to work. It is so that $m^{\phi(n)}\equiv 1\quad(\mod n)$. Who up-voted? – user1537366 Dec 25 '14 at 13:08