Thread: May 2021 View Single Post
2021-05-05, 12:06   #16
Dr Sardonicus

Feb 2017
Nowhere

5×967 Posts

Quote:
Originally Posted by Walter
Quote:
 Given this set, one can generate $A_0$ and $A_1$ that satisfy $A_0$ is equivalent to $F_{m_k-a_k} modulo p_k$ $A_1$ is equivalent to $F_{m_k-a_k+1} modulo p_k$
Should there be a $A_0$ and $A_1$ that satisfies this for every $k$?
Yes. Using the condition that the pk are distinct, the existence of simultaneous solutions to all the congruences is guaranteed by the Chinese Remainder Theorem (CRT).