Fall 2016, problem 24

The Fibonacci sequence is defined by $a_1=a_2=1$ and $a_{k+2}=a_{k+1}+a_k$ for $k\in\mathbb N.$ Show that for any natural number $m$, there exists an index $k$ such that $a_k^4-a_k-2$ is divisible by $m$.


5 years ago

${x^4} - x - 2 = \left( {x + 1} \right)\left( {{x^3} - {x^2} + x - 2} \right)$

if $x \equiv - 1\left[ m \right]$

${x^4} - x - 2$ is a multiple of m

$B = {M_2}\left( \mathbb{Z} \right.$/$\left. {m\mathbb{Z}} \right)$

C is the multiplicative group of invertible elements of B.

C is finite.

$M = \left( {\begin{array}{{20}{c}}\mathop 0\limits^.&\mathop 1\limits^.\\\mathop 1\limits^.&\mathop 1\limits^.\end{array}} \right)$

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

$\left( {\begin{array}{*{20}{c}}{\mathop {{a_k}}\limits^. }\\{\mathop {{a_{k + 1}}}\limits^. }\end{array}} \right)$=${M^k}$ $\left( {\begin{array}{{20}{c}}\mathop 0\limits^.\\\mathop 1\limits^.\end{array}} \right)$=$\left( {\begin{array}{{20}{c}}\mathop 0\limits^.\\\mathop 1\limits^.\end{array}} \right)$

$\mathop {{a_{k - 2}}}\limits^. = - \dot 1$

${a_{k - 2}} \equiv - 1\left[ m \right]$