## Task 30/82

Calculate the product

$$
P_{n}=\prod_{i=0}^{n}\left(2^{2 i}+1\right)
$$
Substituting $n=0 ; 1 ; 2 ; 3$ yields

$$
\begin{array}{r}
p_{0}=3=4-1=2^{2}-1=2^{2^{1}}-1 \\
p_{1}=15=16-1=2^{4}-1=2^{2^{2}}-1 \\
p_{2}=255=256-1=2^{8}-1=2^{2^{3}}-1 \\
p_{3}=65535=65536-1=2^{1} 6-1=2^{2^{4}}-1
\end{array}
$$

This leads to the conjecture that

$$
p_{n}=\prod_{i=0}^{n}\left(2^{2^{i}}+1\right)=2^{2^{n+1}}-1
$$

is true. The conjecture is confirmed (proven to be correct) if it can be shown that the correctness for $n=k$ implies the correctness for $n=k+1$ (principle of mathematical induction). Assuming that

$$
p_{k}=\prod_{i=0}^{k}\left(2^{2^{i}}+1\right)=2^{2^{k+1}}-1
$$

is true, it indeed follows that

$$
p_{k+1}=\prod_{i=0}^{k+1}\left(2^{2^{i}}+1\right)=\prod_{i=0}^{k}\left(2^{2^{i}}+1\right)\left(2^{2^{k+1}}+1\right)=\left(2^{2^{k+1}}-1\right)\left(2^{2^{k+1}}+1\right)=\left(2^{2^{k+2}}-1\right)
$$

Since the base case was established by the substitution, the proof is complete.