Difference between revisions 3684389 and 3684390 on mswiki

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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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This is equivalent to saying that for odd ''n'' all the odd prime factors of F<sub>''n''</sub> are ≡ 1 (mod 4).

<blockquote>For example,
F<sub>1</sub> = 1, F<sub>3</sub> = 2, F<sub>5</sub> = 5, F<sub>7</sub> = 13, F<sub>9</sub> = 34 = 2×17, F<sub>11</sub> = 89, F<sub>13</sub> = 233, F<sub>15</sub> = 610 = 2×5×61
</blockquote>

===Fibonacci 
dand Legendre===

There are some interesting formulas connecting the Fibonacci numbers and the [[dapat beberapa rumus yang menarik menghubungkan nombor Fibonacci dan [[simbol Legendre symbol]] <math>\;\left(\tfrac{p}{5}\right).</math> 

:<math>
\left(\frac{p}{5}\right) 
= \left \{ 
\begin{array}{cl} 0 & \textrm{if}\;p =5
\\ 1 &\textrm{if}\;p \equiv \pm1 \pmod 5
\\ -1 &\textrm{if}\;p \equiv \pm2 \pmod 5
\end{array}
\right.
</math>

IfJika ''p'' is a prime nualah [[nombeor thenperdana]], 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{ and }\;\;\;
 
F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p.
</math> 

<blockquote>
For exampleSebagai 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>

Also, ifJuga, jika ''p'' ≠  5 is an odd primeadalah nuombeor thenperdana 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>
Examples of all the casContoh bagi semua kes:

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

::<math>5F_3^2=20\equiv -1 \pmod {7}\;\;\mbox{ and }\;\;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{ and }\frac{5(\frac{11}{5})- 3}{2}=1.</math>  
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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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