Here's what the Fibonacci sequence modulo m looks like for the first few values
of m.
Fn (mod 2) 1,1,0,1,1,0,1,1,0,1,1,0 ...
Fn (mod 3) 1,1,2,0,2,2,1,0,1,1,2,0,2,2,1. ..
Fn (mod 4) 1,1,2,3,1,0,1,1,2,3,1,0,1,1,2 ...
Fn (mod 5) 1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0, 1, 1,2 ...
Fn (mod 6) 1,1,2,3,5,2,1,3,4,1,5,0,5,5,4,3,1,4,5,3,2,5,1,0,1,1,2,3 ...
Notice in each case that the Fibonacci sequence eventually starts to repeat. In other
words, when we compute the Fibonacci sequence modulo m, we eventually find
two consecutive 1 's appearing, and as soon this happens, the sequence repeats. (We
leave as an exercise for you to prove that this always happens.) Thus there is an
integer N > 1 such that
Fn+N Fn (mod m) for all n = 1, 2, ....
저게시발 퀴즈ㄹㅗ나왔는데 난못풀었음 - DCW
나는 아직 저거 배우려면 멀었는데 역시 잡컴클라스
명컴;;;