Search results
As of 2023, the only known Fermat primes are F0 = 3, F1 = 5, F2 = 17, F3 = 257, and F4 = 65537 (sequence A019434 in the OEIS). The Fermat numbers satisfy the following recurrence relations: for n ≥ 1, for n ≥ 2. Each of these relations can be proved by mathematical induction.
We prove that the Fermat numbers satisfy the following recursion, from which the claimed result will follow: for all , We proceed by induction . For we have , as desired.
24 sty 2015 · The Fermat numbers are defined by $F_m = 2^{2^m} + 1$. Prove that for $m \ne n$ we have $(F_m, F_n) = 1$. I have to first prove that $F_{m+1} = F_0F_1 \cdots F_m + 2$ by representing $F_{m+1}$ in terms of $F_m$.
A Fermat number Fn = 2 6 Ù+ 1 (for n ≥ 1) can be thought of as a square whose side length is 2 6 Ù 7 - plus a unit square (see figure1). Hence, determining whether a (Fermat) number is a composite or not is equivalent to determining whether we can rearrange the unit-square blocks to form a rectangle (see figure2).
Proof. Let m and n be distinct Fermat numbers with m > n; and suppose that d > 0 is a common divisor of m and n; then d divides 2 = m 0 1 n m 1: Therefore, d = 1 or d = 2; but m and n are odd, so we must have d = 1: Therefore, for m > n; the Fermat numbers m and n are relatively prime.
First we present a few recurrence formulae for the Fermat numbers. Most of these can be found in the paper [Grytczuk] (see also [Schram]). Proposition 3.1. The formula. holds for all m 2: 1. Fro 0 f . We have. By induction, we get easily that for all k E {I, ... , m}. Now formula (3.1) is an immediate consequence of formulae (3.2) and (3.3).
He presented 12 theorems with proof including the binomial theorem for positive exponents and a combinatorial proof of the recursive formula for forming the triangle. He also showed how to apply the triangle by using it to answer an outstanding problem on how to divide a bet when a game of chance had been stopped before a winner was determined.
