# Product of Sequence of Fermat Numbers plus 2

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## Contents

## Theorem

Let $F_n$ denote the $n$th Fermat number.

Then:

\(\displaystyle \forall n \in \Z_{>0}: \ \ \) | \(\displaystyle F_n\) | \(=\) | \(\displaystyle \prod_{j \mathop = 0}^{n - 1} F_j + 2\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F_0 F_1 \dotsm F_{n - 1} + 2\) |

### Corollary

Let $F_n$ denote the $n$th Fermat number.

Let $m \in \Z_{>0}$ be a (strictly) positive integer.

Then:

- $F_n \divides F_{n + m} - 2$

where $\divides$ denotes divisibility.

## Proof

The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

- $F_n = \displaystyle \prod_{j \mathop = 0}^{n - 1} F_j + 2$

### Basis for the Induction

$\map P 1$ is the case:

\(\displaystyle F_1\) | \(=\) | \(\displaystyle 2^{\paren {2^1} } + 1\) | Definition of Fermat Number | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 5\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 3 + 2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {2^{\paren {2^0} } + 1} + 2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F_0 + 2\) | Definition of Fermat Number | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \displaystyle \prod_{j \mathop = 0}^{1 - 1} F_j + 2\) |

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

- $F_k = \displaystyle \prod_{j \mathop = 0}^{k - 1} F_j + 2$

from which it is to be shown that:

- $F_{k + 1} = \displaystyle \prod_{j \mathop = 0}^k F_j + 2$

### Induction Step

This is the induction step:

\(\displaystyle F_{k + 1}\) | \(=\) | \(\displaystyle 2^{\paren {2^{k + 1} } } + 1\) | Definition of Fermat Number | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{\paren {2 \times 2^k} } + 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2^{\paren {2^k} } \times 2^{\paren {2^k} } + 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {F_k - 1} \times \paren {F_k - 1} + 1\) | Definition of Fermat Number | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\paren {\prod_{j \mathop = 0}^{k - 1} F_j + 2} - 1} \paren {F_k - 1} + 1\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\prod_{j \mathop = 0}^{k - 1} F_j} F_k + F_k - \prod_{j \mathop = 0}^{k - 1} F_j - 1 + 1\) | multiplying out | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \prod_{j \mathop = 0}^k F_j + F_k - \paren {F_k - 2}\) | Induction Hypothesis, and simplifying | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \prod_{j \mathop = 0}^k F_j + 2\) | simplifying |

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

- $\forall n \in \Z_{>0}: F_n = \displaystyle \prod_{j \mathop = 0}^{n - 1} F_j + 2$

$\blacksquare$

## Sources

- 1986: David Wells:
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