Let $p$ be a prime.
Consider the recurrence equation $$s_{n}=(s_{n-1}²-2)(mod(2^{p}-1))$$ where $s₀=4$
My question is: Can we write this recurrence as follow? $$s_{n}=(2^{p}-1)q+(s_{n-1}²-2)$$ where $q$ is an integer.
Asked
Active
Viewed 90 times
1 Answers
1
This follow by definition of congruency if $x\equiv b(mod\ c)$ then$x=b+c\cdot q$ where q is an integer.Substitute b=$s_{n-1}^2-2$ ,$c=2^p-1$,x=$s_{n}$ .
mcmat23
- 1,050
-
@ mcmat23: Is $q$ in the question should be $q_{n}$ since we make the devision of $s_{n}$ and $2^{p}-1$. – DER Oct 14 '14 at 17:49
-
Yes ,q changes for each terms. – mcmat23 Oct 14 '14 at 18:45