Questions surface:
text:
The combination of Equations title together becomes:
\(X^{a+c}\equiv b \cdot d (\mod p)\)
That time, we assume that two numbers \ (x \) and \ (the y-\) , such that:
\(ax + cy = 1\)
then:
\(X^{ax+cy}\equiv X \equiv b^x \cdot d^y (\mod p)\)
Then we can \ (ax + cy = 1 \ ) run again expanding in Europe, then according to \ (X-\ equiv B X ^ \ Y ^ CDOT D (\ MOD P) \) , able to come \ (X-\ ) a.
But do you think the topic of people so good it?
\ (X \) and \ (Y \) may be negative, do \ (b ^ x \ cdot d ^ y \) is equivalent to the time \ (\ frac {1} { b ^ {(- x)}} \ CDOT \ FRAC. 1 {{} {D ^ (- Y)}} \) , because the same membrane skill I, certainly pan here.
So we have to \ (b \) and \ (d \) to pray for inverse, meanwhile, will also expand with Europe.