Consider the sequence \(x_k\) of positive numbers given by \(a_k=\cosh x_k\) (\(\cosh\) is the hyperbolic cosine, defined by \(\cosh t=\frac{e^t+e^{-t}}2\)). It is easily verified that \(\cosh(2x_k)=2a_k^2-1=\cosh x_{k+1}\), so \(x_{k+1}=2x_k\), i.e. \(x_k=\lambda\cdot 2^k\) for some \(\lambda > 0\). The condition \(a_0=2\) gives us \(\lambda=\log(2+\sqrt3)\). Therefore \[a_n=\frac{(2+\sqrt3)^{2^n}+(2-\sqrt3)^{2^n}}2.\] Let \(p > 2\) be a prime number such \(p\mid a_n\). We distinguish two cases.
- \(1^{\circ}\) \(m^2\equiv 3\) (mod \(p\)) for some \(m\in\mathbb{Z}\). Then \[(2+m)^{2^n}+(2-m)^{2^n}\equiv 0\;(\mbox{mod }p).\quad\quad\quad\quad\quad (1)\] Since \((2+m)(2-m)=4-m^2\equiv 1\) (mod \(p\)), multiplying both sides of (1) by \((2+m)^{2^n}\) yields \((2+m)^{2^{n+1}}\equiv-1\) (mod \(p\)). It follows that the order of number \((2+m)\) modulo \(p\) equals \(2^{n+2}\), from which we conclude \(2^{n+2}\mid p-1\).
- \(2^{\circ}\) The congruence \(m^2\equiv 3\) (mod \(p\)) has no solutions. We work in the quadratic extension \(\mathbb{Z}_p(\sqrt3)\) of the field \(\mathbb{Z}_p\) in which number \(\sqrt3\) actually plays the role of the number \(m\) from case (1). As in case (1) we have \((2+\sqrt3)^{2^{n+1}}=-1\), which means that the order of \(2+\sqrt3\) in the multiplicative group \(\mathbb{Z}_p(\sqrt3)^\ast\) equals \(2^{n+2}\). This is not enough to finish the proof as in case (1), as the group \(\mathbb{Z}_p(\sqrt3) ^\ast\) has \(p^2-1\) elements; instead, we only get that \(2^{n+2}\mid p^2-1\). However, we shall be done if we find \(u\in\mathbb{Z}_p(\sqrt3)\) for which \(u^2=2+\sqrt3\): indeed, then the order of \(u\) is \(2^{n+3}\), so \(2^{n+3}\mid p^2-1\) and therefore \(2^{n+2}\mid p-1\), since \(4\nmid p+1\).
Note that \((1+\sqrt3)^2=2(2+\sqrt3)\). Now it is enough to show that \(1/2\) is a perfect square in the field \(\mathbb{Z}_p(\sqrt3)\). This immediately follows from the relation \(a_n=0=2a_{n-1}^2-1\), as \(1/2=a_{n-1}^2\). This finishes the proof.