Product of Sequence of Fermat Numbers plus 2

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Theorem

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


Then:

\(\ds \forall n \in \Z_{>0}: \, \) \(\ds F_n\) \(=\) \(\ds \prod_{j \mathop = 0}^{n - 1} F_j + 2\)
\(\ds \) \(=\) \(\ds F_0 F_1 \dotsm F_{n - 1} + 2\)


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:

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


Basis for the Induction

$\map P 1$ is the case:

\(\ds F_1\) \(=\) \(\ds 2^{\paren {2^1} } + 1\) Definition of Fermat Number
\(\ds \) \(=\) \(\ds 5\)
\(\ds \) \(=\) \(\ds 3 + 2\)
\(\ds \) \(=\) \(\ds \paren {2^{\paren {2^0} } + 1} + 2\)
\(\ds \) \(=\) \(\ds F_0 + 2\) Definition of Fermat Number
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 0}^{1 - 1} F_j + 2\)

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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

\(\ds F_{k + 1}\) \(=\) \(\ds 2^{\paren {2^{k + 1} } } + 1\) Definition of Fermat Number
\(\ds \) \(=\) \(\ds 2^{\paren {2 \times 2^k} } + 1\)
\(\ds \) \(=\) \(\ds 2^{\paren {2^k} } \times 2^{\paren {2^k} } + 1\)
\(\ds \) \(=\) \(\ds \paren {F_k - 1} \times \paren {F_k - 1} + 1\) Definition of Fermat Number
\(\ds \) \(=\) \(\ds \paren {\paren {\prod_{j \mathop = 0}^{k - 1} F_j + 2} - 1} \paren {F_k - 1} + 1\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {\prod_{j \mathop = 0}^{k - 1} F_j} F_k + F_k - \prod_{j \mathop = 0}^{k - 1} F_j - 1 + 1\) multiplying out
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 0}^k F_j + F_k - \paren {F_k - 2}\) Induction Hypothesis, and simplifying
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 0}^k F_j + 2\) simplifying

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


Therefore:

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

$\blacksquare$


Sources

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