RSA cryptography

Question

I saw on Wikipedia RSA algorithm and the private key has a condition imposed on it which says $$d \equiv e^{-1} \mod \phi(n)$$ where $n =(p-1)(q-1)$ but after a few steps $d$ condition becomes $$de \equiv 1 \mod \phi(n).$$ Do both of these mean the same?

Answer 1

Yes, they mean the same thing. By definition, $x^{-1}x = 1 \bmod{\phi(n)}$ for all $x$ in the group. So we can do

$$ d=e^{-1}\bmod{\phi(n)}$$ $$ de=e^{-1}e\bmod{\phi(n)}$$ $$ de=1\bmod{\phi(n)}$$

We know that $e$ is in the group, because $gcd(e,\phi(n))=1$, so the inverse exists.