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Let us define  $$a+b\sqrt{5} \equiv c+d\sqrt{5} \pmod{p}$$ as $a \equiv 0 \pmod{p}$, $b \equiv 0 \pmod{p}$. 
Note for a prime $p$ we may say $$(1+\sqrt{5})^{p} \equiv \sum_{k=0}^{p}\left(\sqrt{5}\right)^{k}\binom{p}{k} \equiv 1+(\sqrt{5})^{p} \equiv 1+\sqrt{5}\left(\frac{5}{p}\right) \pmod{p} \tag{1}$$
And 
$$(1-\sqrt{5})^{p} \equiv \sum_{k=0}^{p}\left(-\sqrt{5}\right)^{k}\binom{p}{k} \equiv 1+(-\sqrt{5})^{p} \equiv 1-\sqrt{5}\left(\frac{5}{p}\right) \pmod{p}\tag{2}$$
If $p \equiv 1,4 \pmod{5}$, if we multiply each side of $(1)$ by $\sqrt{5}-1$. $$4 \times (\sqrt{5}+1)^{p-1} \equiv 4 \pmod{p}$$
 And so $$(\sqrt{5}+1)^{p-1} \equiv 1 \pmod{p}$$
In the same way, we may say $$(1-\sqrt{5})^{p-1} \equiv 1 \pmod{p}$$
And, if $p \equiv 2,3 \pmod{5}$, we may say $$(1+\sqrt{5})^{p+1} \equiv -4 \pmod{p}$$ and $$(1-\sqrt{5})^{p+1} \equiv -4 \pmod{p}$$
Thus we may say that for all odd primes $p \neq 5$
 $$\frac{1}{2^{p -\left(\frac{5}{p}\right)}\sqrt{5}}\left((1 +\sqrt{5})^{p -\left(\frac{5}{p}\right)}-(1-\sqrt{5})^{p -\left(\frac{5}{p}\right)}\right) \equiv 0 \pmod{p}$$
So $$F_{p -\left(\frac{5}{p}\right)} \equiv 0 \pmod{p}$$
if $p$ is a odd prime number not dividisble by $5$.

Thu, 05 Oct 2017 11:59 GMT