Content deleted Content added
mNo edit summary |
mNo edit summary |
||
Line 13:
& = & (g^{\phi(p)/p_1})^{b_1} \pmod{p}
\end{matrix}
</math> (using [[Euler's theorem]])</center><br> The same operation is now performed for ''p<sub>2</sub>'' up to ''p<sub>n</sub>''.<br><br>A minor modification is needed where a prime number is repeated. Suppose we are seeing ''p<sub>i</sub>'' for the ''k+1''-th time. Then we already know ''c<sub>i</sub>'' in the equation ''x = a<sub>i</sub> p<sub>i</sub><sup>k+1</sup> + b<sub>i</sub> p<sub>i</sub><sup>k</sup>+c<sub>i</sub>'', and we find ''b<sub>i</sub>'' the same way as before.
#We end up with enough simultaneous congruencies so that ''x'' can be solved using the [[Chinese remainder theorem]].
| |||