Let \((a_n)\) be a sequence of real numbers satisfying \(a_0=1\) and \(a_{n+1}=a_0a_1\cdots a_n+4\) for all natural numbers \(n\). Prove that \(a_n-\sqrt{a_{n+1}}=2\) for all \(n>0\).

We observe that \(a_k>0\) for all natural numbers \(k\).

From \(a_{n+1}=a_0a_1\cdots a_n+4\), replacing \(n\) by \(n+1\), we get:

\(a_{n+2}=a_0a_1\cdots a_{n+1}+4=(a_0a_1\cdots a_n)(a_0a_1\cdots a_n+4)+4\)

\(a_{n+2}=(a_0a_1\cdots a_n)^2+4(a_0a_1\cdots a_n)+4=(a_0a_1\cdots a_n+2)^2\)

\(\sqrt{a_{n+2}}=a_0a_1\cdots a_n+2=a_{n+1}+2\)

\(\sqrt{a_{n+1}}=a_n+2\)

Therefore, \(a_n-\sqrt{a_{n+1}}=a_n-(a_n+2)=-2\).