Show that there exists a Fibonacci number that is divisible by 2014.i s this as simple as showing an nth F(n) is divisible by 2? A prime factor of 2014 . Some reason tonight my math faculties only at 40% :)
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Related : (http://math.stackexchange.com/q/1266156) – Jean Marie Aug 05 '16 at 04:30
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$F(0)=0$ is a multiple of $2014$.
Suppose we exclude $0$, it is still true. Lets take the first two values $n<m$ such that $F(n)\equiv F(m)\bmod 2014$ and $F(n+1)\equiv F(m+1)\bmod 2014$ (they must exist since there are only $2014^2$ options for $(F(n),F(n+1))\bmod 2014$ ).
We must have $n=0$, otherwise notice $F(n-1)\equiv F(n+1)-F(n)\equiv F(m+1)-F(m)\equiv F(m-1)$, contradicting the minimality of $n$.
Therefore $F(m)\equiv F(0)\equiv 0\bmod 2014$. We are done.
Asinomás
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@RandinMichaelDivelbiss You may want to know that generally for any $n$, there is such a $m$ that $F_{m}$ is divisible by $n$. However, finding such a $m$ is not always easy. – S.C.B. Aug 05 '16 at 03:52
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@S.C.B., no, although if $n$ is squarefree we can do some magic with matrices. – Asinomás Aug 05 '16 at 03:52
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@CarryonSmiling You mean we can't do it for all $n$? My book says that you can. – S.C.B. Aug 05 '16 at 03:53
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@S.C.B. I meant that if $n$ is square free we can find an explicit $k$ such that $n|F(k)$, using matrices. – Asinomás Aug 05 '16 at 03:54
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