Undergrad A question about a small step in the proof of RSA encryption

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The discussion centers on a specific step in the RSA encryption proof, particularly focusing on the implications of equation 5 from the referenced paper. It clarifies that the relationship \( ed \equiv 1 \mod \phi(n) \) leads to the conclusion that \( \phi(n) \) divides \( ed - 1 \). This division implies that there exists an integer \( k \) such that \( ed \) can be expressed as \( \phi(n) \cdot k + 1 \). Consequently, this formulation allows for the equivalence \( M^{\phi(n) \cdot k + 1} = M^{ed} \). Understanding this step is crucial for grasping the foundational principles of RSA encryption.
Leo Liu
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From the paper https://people.csail.mit.edu/rivest/Rsapaper.pdf
Can someone explain the green highlight to me please? Sorry that I can't type much because this is the final week. Thanks.
 
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It should be evident from equation 5
 
From ##(5)## we have
\begin{align*}
ed\equiv 1 \mod \phi(n) &\Longleftrightarrow \phi(n)\,|\,(ed-1) \\
&\Longleftrightarrow \phi(n)\cdot k = ed-1 \text{ for some } k \in \mathbb{Z}\\
&\Longleftrightarrow \phi(n)\cdot k +1 = ed \text{ for some } k \in \mathbb{Z}\\
&\Longrightarrow M^{\phi(n)\cdot k +1} =M^{ed}
\end{align*}
 
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Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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