中国剩余定理证明
\[dp = d \mod {p-1}
\\dq = d \mod {q-1}
\\e = 65537
\\CRT
\left \{ \begin{array}{c}
x \equiv a_1 \mod n_1
\\x \equiv a_2 \mod n_2
\\...\end{array}
\right .
\\n_1,n_2,...,n_i两两互素
\\x一定存在，且存在构造法可解
\\令M = \prod_{1}^{k}{n_i}
\\有M_i = \frac{M}{n_i}
\\设M_i' 使得，M_i'* M_i \equiv 1 mod n_i
\\则x = \sum_{1}^{i}a_i * M_i * M_i' \mod {M}
\\ (a+b) \mod n = a \mod n + b \mod n
\\ 证x\equiv a_k \mod n_k
\\ 当i=k:
\\(a_k * M_k * M_k' \mod M) \mod n_k
\\= a_k * M_k * M_k' \mod n_k
\\= a_k \mod n_k
\\
\\当i \ne k:
\\(a_i * M_i * M_i' \mod M) \mod n_k
\\a_i * \frac{M}{n_i} * M_i' \mod n_k
\\=0
\\\
\\d = dp \mod {p-1}
\\d = dq \mod {q-1}
\\令g = gcd(p-1,q-1)
\\p-1 = g*k_p,q-1 = g *k_q
\\\left \{ \begin{array} {c}
d \equiv d_1 \mod g
\\ d \equiv d_2 \mod k_p
\\ d \equiv d_3 \mod k_q
\end{array}
\right .
\\ d \equiv dp \mod g
\\dp = d \mod {p-1}
\\dp = d \mod {k_q*g}
\\(d \mod {k_q*g}) \equiv dp \mod p-1
\\\
\\\left \{ \begin{array} {c}
d \equiv dp \mod g
\\ d \equiv dp \mod k_p
\\ d \equiv dq \mod k_q
\end{array}
\right .
\]
