Thread: May 2021
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Old 2021-05-05, 12:06   #16
Dr Sardonicus
 
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Quote:
Originally Posted by Walter View Post
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).
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