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Fall 2016, problem 24

The Fibonacci sequence is defined by a1=a2=1 and ak+2=ak+1+ak for kN. Show that for any natural number m, there exists an index k such that a4kak2 is divisible by m.

Comments

francois
5 years ago

x4x2=(x+1)(x3x2+x2)

if x1[m]

x4x2 is a multiple of m

B=M2(Z/mZ)

C is the multiplicative group of invertible elements of B.

C is finite.

M=(.0.1.1.1)

is an invertible element of B. M has a multiplicative order k.

(.ak.ak+1)=Mk (.0.1)=(.0.1)

.ak2=˙1

ak21[m]