I am preparing for the final exam in my Data Security class so I am trying to read and understand the textbook's exercise questions. The sample solution at the end the textbook is more like a hint instead of being a step-by-step solution.
Question:
Textbook solution: Yes. They are equivalent.
My answer: I do not know to check if some scheme is equivalent to RSA. So I tried to verify $D$.
Step 2: $(P-1)(Q-1) - 1 = (PQ -P-Q+1) - 1 = (N -P-Q+1) - 1 = (N-(P+Q-1) - 1 \equiv 0 \ (mod \ E)$
Step 4: $DE = (P-1) (Q-1) (E-1) + 1 = (PQ - P-Q + 1) (E-1) + 1 = (N - P-Q + 1) (E-1) + 1 = NE-N-E(P+Q-1) + 1$
Take $(mod \ E)$ of both sides:
$[DE = NE-N-E(P+Q-1) + 1] \ (mod \ E)$
$0 \ (mod \ E) \equiv 0 - N - 0 + (P+Q-1) + 1 \ (mod \ E)$
$0 \ (mod \ E) \equiv - (N - (P+Q-1)) - 1 \ (mod \ E)$
$0 \ (mod \ E) \equiv 0 \ (mod \ E) \leftarrow$ verified $D$ is correct.
But obviously this is not a concrete reasoning on why this scheme is equivalent to RSA. Any hints would be appreciated.