这是一份oxford牛津大学作业代写的成功案例
Unfortumately, the answer is yes. The smallest such number is $341=11 \cdot 31$. This is not a prime, but it satisfies
$$
341 \mid 2^{340}-1 .
$$
(How do we know that this divisibility relation holds without extensive computation? We can use Fermat’s Theorem. It is sufficient to argue that both 11 and 31 are divisors of $2^{340}-1$, since then so is their product, 11 and 31 being different primes. By Fermat’s Theorem,
$$
11 \mid 2^{10}-1
$$
Next we invoke the result : It implies that
$$
2^{10}-1 \mid 2^{340}-1
$$
Hence
$$
11 \mid 2^{340}-1
$$
For 31, we don’t need Fermat’s Theorem, but only again:
$$
31=2^{5}-1 \mid 2^{340}-1
$$
Oxford COURSE NOTES ：
4.3.2. For $n=1$ and $n=2$, if we require that $L_{n}$ be of the given form, then we get
$$
L_{1}=1=a+b, \quad L_{2}=3=a \frac{1+\sqrt{5}}{2}+b \frac{1-\sqrt{5}}{2}
$$
Solving for $a$ and $b$, we get
$$
a=\frac{1+\sqrt{5}}{2}, \quad b=\frac{1-\sqrt{5}}{2}
$$
Then
$$
L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}
$$