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#From the remainder theorem we know that ''x = a<sub>1</sub> p<sub>1</sub> + b<sub>1</sub>''. We now find the value of ''b<sub>1</sub>'' for which the following equation holds using a fast algorithm such as [[Baby-step giant-step]] :<br> <center><math>
\begin{matrix}e^{\phi(p)/p_1} & = & (g^x)^{\phi(p)/p_1} \pmod{p} \\
& = & (g^{\phi(p)})^{a_1}g^{b_1\phi(p)/p_1} \pmod{p} \\
& = & (g^{\phi(p)/p_1})^{b_1} \pmod{p}
\end{matrix}
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