Difference between revisions 3684390 and 3684391 on mswiki

{{pelbagai isu|{{cleanup|reason=memerlukan penterjemahan segera kerana sudah ditinggalkan sejak tahun 2008|date=Ogos 2014}}{{Terjemah|en|fabonacci number|date=Ogos 2014}}}}
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[[Image:FibonacciBlocks.svg|thumb|180px|right|Suatu ubinan dengan segi empat yang tepinya adalah nombor Fibonaci berturut-turut pada panjangnya]]
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\end{array}
\right.
</math>

Jika ''p'' ialah [[nombor perdana]], maka<ref>[[Paulo Ribenboim]] (1996), ''The New Book of Prime Number Records'', New York: Springer, ISBN 0-387-94457-5, p. 64</ref><ref>Franz Lemmermeyer (2000), ''Reciprocity Laws'', New York: Springer, ISBN 3-540-66957-4, ex 2.25-2.28, pp. 73-74</ref>
<math>
F_{p} \equiv \left(\frac{p}{5}\right) \pmod p \;\;\mbox{ 
dand }\;\;\;
 
F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p.
</math> 

<blockquote>
Sebagai contoh, 

:<math>(\tfrac{2}{5}) = -1, \,\, F_3  = 2, F_2=1,</math> 
:<math>(\tfrac{3}{5}) = -1, \,\, F_4  = 3,F_3=2,</math> 
:<math>(\tfrac{5}{5}) = \;\;\,0,\,\,  F_5  = 5,</math> 
:<math>(\tfrac{7}{5}) = -1,  \,\,F_8  = 21,\;\;F_7=13,</math> 
:<math>(\tfrac{11}{5}) = +1,  F_{10}  = 55, F_{11}=89.</math> 

</blockquote>

Juga, jika ''p'' ≠  5 adalah nombor perdana ganjil, maka: <ref>Lemmermeyer, ex. 2.38, pp. 73-74</ref>
:<math>5F^2_{\left(p \pm 1 \right) / 2}
\equiv
\begin{cases} 
\frac{5\left(\frac{p}{5}\right)\pm 5}{2} \pmod p & \textrm{if}\;p \equiv 1 \pmod 4\\
\\
\frac{5\left(\frac{p}{5}\right)\mp 3}{2} \pmod p & \textrm{if}\;p \equiv 3 \pmod 4
\end{cases}
</math>

<blockquote>
Contoh bagi semua kes:

:<math>p=7 \equiv 3 \pmod 4, \;\;(\tfrac{7}{5}) = -1, \frac{5(\frac{7}{5})+3}{2} =-1\mbox{ dand }\frac{5(\frac{7}{5})-3}{2}=-4.</math>  
::<math>F_3=2 \mbox{ dand } F_4=3.</math> 

::<math>5F_3^2=20\equiv -1 \pmod {7}\;\;\mbox{ dand }\;\;5F_4^2=45\equiv -4 \pmod {7}</math>

:<math>p=11 \equiv 3 \pmod 4, \;\;(\tfrac{11}{5}) = +1, \frac{5(\frac{11}{5})+3}{2} =4\mbox{ dand }\frac{5(\frac{11}{5})- 3}{2}=1.</math>  
::<math>F_5=5 \mbox{ dand } F_6=8.</math> 

::<math>5F_5^2=125\equiv 4 \pmod {11} \;\;\mbox{ dand }\;\;5F_6^2=320\equiv 1 \pmod {11}</math>

:<math>p=13 \equiv 1 \pmod 4, \;\;(\tfrac{13}{5}) = -1, \frac{5(\frac{13}{5})-5}{2} =-5\mbox{ dand }\frac{5(\frac{13}{5})+ 5}{2}=0.</math>  
::<math>F_6=8 \mbox{ dand } F_7=13.</math> 

::<math>5F_6^2=320\equiv -5 \pmod {13} \;\;\mbox{ dand }\;\;5F_7^2=845\equiv 0 \pmod {13}</math>

:<math>p=29 \equiv 1 \pmod 4, \;\;(\tfrac{29}{5}) = +1, \frac{5(\frac{29}{5})-5}{2} =0\mbox{ dand }\frac{5(\frac{29}{5})+5}{2}=5.</math>  
::<math>F_{14}=377 \mbox{ dand } F_{15}=610.</math> 

::<math>5F_{14}^2=710645\equiv 0 \pmod {29} \;\;\mbox{ dand }\;\;5F_{15}^2=1860500\equiv 5 \pmod {29}</math>
</blockquote>

===Kebolehbahagian dengan 11===
<math>\sum_{k=n}^{n+9} F_{k} = 11 F_{n+6}</math>

<blockquote>Sebagai contoh, jadi ''n'' = 1:
<br>
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*[http://web.archive.org/web/20070715032716/http://mathdl.maa.org/convergence/1/?pa=content&sa=viewDocument&nodeId=630&bodyId=1002 Fibonacci Numbers] at [http://web.archive.org/web/20060212072618/http://mathdl.maa.org/convergence/1/ Convergence]
* [http://www.tools4noobs.com/online_tools/fibonacci/ Online Fibonacci calculator]

[[Kategori:Fibonacci numbers|*]]
[[Kategori:Articles containing proofs]]

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