## Task $7 / 71$

Let $p_{n}$ be the n-th prime number. Find all $i$ such that $p_{i}=2 i+1$.
Let

\[
\begin{aligned}
& p_{1}=2 \neq 2 \cdot 1+1, \quad p_{2}=3 \neq 2 \cdot 2+1, \quad p_{3}=5 \neq 2 \cdot 3+1 \\
& p_{4}=7 \neq 2 \cdot 4+1, \quad p_{5}=11=2 \cdot 5+1, \quad p_{6}=13=2 \cdot 6+1, \quad p_{7}=17 \neq 2 \cdot 7+1
\end{aligned}
\]

We claim that \(i=5\) and \(i=6\) are the only \(i\) for which \(p_{i}=2 i+1\). Indeed, for \(k \geq 0\), it holds that

\[
p_{7+k} \geq 17+2 k=2(7+k)+3>2(7+k)+1
\]