Checking whether a number is prime or not 
Is it true that a natural number $n>1$ is prime if and only if $n|\left ( \frac{1+\sqrt{5}}{2} \right )^n+\left ( \frac{1-\sqrt{5}}{2} \right )^n-1$?

We know that $11$ is a prime number, but let us assume that we do not know, and let us also assume that the statement is true;
$$\left ( \frac{1+\sqrt{5}}{2} \right )^{11}+\left ( \frac{1-\sqrt{5}}{2} \right )^{11}-1=198.$$
Clearly, $11|198$. Therefore, as our assumption that the statement is true, the number $11$ is a prime number.

Note that $\left ( \frac{1+\sqrt{5}}{2} \right )^n+\left ( \frac{1-\sqrt{5}}{2} \right )^n$ is the $n$-th Lucas number.
It is true that if $p$ is a prime then $L_p-1$ is divisible by $p$: $2$ divides $L_2-1=2$ and for any prime $p>2$,
$$L_p-1=\frac{1}{2^{p-1}}\sum_{k=1}^{(p-1)/2}\binom{p}{2k}5^k=0\pmod{p}$$
because $p$ divides each $\binom{p}{2k}$. 
On the contrary $705$ is not a prime but it divides $L_{705}-1$. 
See Bruckman-Lucas_pseudoprimes and compare with the Wall-Sun-Sun prime.
