I have found two ways to compute the RSA private key, but i have trouble understanding why they are equivalent.

$1) D={k \phi(N)+1 \over E}$

$2) E* D=1 \mod \phi(N)$

$1)$ AFAIK comes from taking the relationship between eulers theorem and a modular identity such that. $$ M^{1+ \phi(N)} \equiv M \mod N \qquad \land \qquad M^{E*D} \equiv M \mod N \\ \qquad \text{ Euler} \qquad \qquad \qquad \qquad \qquad \qquad \text{ Identity}$$

So we can see that the exponents must be the same, and that exponentiating by an integer $K$ should make no difference in modular arithmethic

$k \phi(N)+1 = E*D$


$D={k \phi(N)+1 \over E}\\$

The origin of $2)$ eludes me, why is it equivalent to $1)$?


$D={k \phi(N)+1 \over E}$ is equivalent to $ED = k \phi(N) + 1$

$ED \equiv 1 \pmod {\phi(N)}$ means "there's some integer $k$ s.t. $ED = 1 + k \phi(N)$"

Hence, they're equivalent.

BTW: the actual necessary and sufficient condition on E, D actually is $ED \equiv 1 \pmod {\text{lcm}(P-1, Q-1)}$ (for RSA modulii that consists of two primes); the relation you're using is sufficient, but not necessary.


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