In mathematics, a Fermat number, named after Pierre de Fermat who first studied them, is a positive integer of the form \[ F_{n} = 2^{2^{ \overset{n} {}}} + 1 \] where n is a nonnegative integer. The first few Fermat numbers are: 3, 5, 17, 257, 65537, 4294967297, 18446744073709551617, … (sequence A000215 in OEIS). If 2n + 1 is prime, and n > 0, it can be shown that n must be a power of two. (If n = ab where 1 ≤ a, b ≤ n and b is odd, then 2n + 1 = (2a)b + 1 ≡ (−1)b + 1 = 0 (mod 2a + 1). See below for complete proof.) In other words, every prime of the form 2n + 1 is a Fermat number, and such primes are called Fermat primes. The only known Fermat primes are F0, F1, F2, F3, and F4. Basic properties The Fermat numbers satisfy the following recurrence relations: \[ F_{n} = (F_{n1}1)^{2}+1\! \] for n ≥ 1, \[ F_{n} = F_{n1} + 2^{2^{n1}}F_{0} \cdots F_{n2}\! \] \[ F_{n} = F_{n1}^2  2(F_{n2}1)^2\! \] \[ F_{n} = F_{0} \cdots F_{n1} + 2\! \] for n ≥ 2. Each of these relations can be proved by mathematical induction. From the last equation, we can deduce Goldbach's theorem: no two Fermat numbers share a common factor. To see this, suppose that 0 ≤ i < j and Fi and Fj have a common factor a > 1. Then a divides both \[ F_{0} \cdots F_{j1} \] and Fj; hence a divides their difference, 2. Since a > 1, this forces a = 2. This is a contradiction, because each Fermat number is clearly odd. As a corollary, we obtain another proof of the infinitude of the prime numbers: for each Fn, choose a prime factor pn; then the sequence {pn} is an infinite sequence of distinct primes. Further properties: The number of digits D(n,b) of Fn expressed in the base b is \[ D(n,b) = \left\lfloor \log_{b}\left(2^{2^{\overset{n}{}}}+1\right)+1 \right\rfloor \approx \lfloor 2^{n}\,\log_{b}2+1 \rfloor \] (See floor function). No Fermat number can be expressed as the sum of two primes, with the exception of F1 = 2 + 3. Primality of Fermat numbers Fermat numbers and Fermat primes were first studied by Pierre de Fermat, who conjectured (but admitted he could not prove) that all Fermat numbers are prime. Indeed, the first five Fermat numbers F0,...,F4 are easily shown to be prime. However, this conjecture was refuted by Leonhard Euler in 1732 when he showed that \[ F_{5} = 2^{2^5} + 1 = 2^{32} + 1 = 4294967297 = 641 \cdot 6700417. \; \] Euler proved that every factor of Fn must have the form k2n+2 + 1. The fact that 641 is a factor of F5 can be easily deduced from the equalities 641 = 27×5+1 and 641 = 24 + 54. It follows from the first equality that 27×5 ≡ −1 (mod 641) and therefore (raising to the fourth power) that 228×54 ≡ 1 (mod 641). On the other hand, the second equality implies that 54 ≡ −24 (mod 641). These congruences imply that −232 ≡ 1 (mod 641). It is widely believed that Fermat was aware of the form of the factors later proved by Euler, so it seems curious why he failed to follow through on the straightforward calculation to find the factor.[1] One common explanation is that Fermat made a computational mistake and was so convinced of the correctness of his claim that he failed to doublecheck his work. There are no other known Fermat primes Fn with n > 4. However, little is known about Fermat numbers with large n.[2] In fact, each of the following is an open problem: Is Fn composite for all n > 4? As of 2010 it is known that Fn is composite for 5 ≤ n ≤ 32, although complete factorizations of Fn are known only for 0 ≤ n ≤ 11, and there are no known factors for n in {20, 24}.[4] The largest Fermat number known to be composite is F2543548, and its prime factor 9×22543551 + 1 was discovered by Scott Brown in PrimeGrid's Proth Prime Search on June 22, 2011. The following heuristic argument suggests there are only finitely many Fermat primes: according to the prime number theorem, the "probability" that a number n is prime is at most A/ln(n), where A is a fixed constant. Therefore, the total expected number of Fermat primes is at most \[ \begin{align}A \sum_{n=0}^{\infty} \frac{1}{\ln F_{n}} &= \frac{A}{\ln 2} \sum_{n=0}^{\infty} \frac{1}{\log_{2}(2^{2^{n}}+1)}\\ &< \frac{A}{\ln 2} \sum_{n=0}^{\infty} 2^{n} \\ &= \frac{2A}{\ln 2}.\end{align} \] It should be stressed that this argument is in no way a rigorous proof. For one thing, the argument assumes that Fermat numbers behave "randomly", yet we have already seen that the factors of Fermat numbers have special properties. If (more sophisticatedly) we regard the conditional probability that n is prime, given that we know all its prime factors exceed B, as at most Aln(B)/ln(n), then using Euler's theorem that the least prime factor of Fn exceeds 2n + 1, we would find instead \[ \begin{align}A \sum_{n=0}^{\infty} \frac{\ln 2^{n+1}}{\ln F_{n}} &= A \sum_{n=0}^{\infty} \frac{\log_2 2^{n+1}}{\log_{2}(2^{2^{n}}+1)} \\ &< A \sum_{n=0}^{\infty} (n+1) 2^{n} \\ &= 4A.\end{align} \] Although such arguments engender the belief that there are only finitely many Fermat primes, one can also produce arguments for the opposite conclusion. Suppose we regard the conditional probability that n is prime, given that we know all its prime factors are 1 modulo M, as at least CM/ln(n). Then using Euler's result that M = 2n + 1 we would find that the expected total number of Fermat primes was at least \[ \begin{align}C \sum_{n=0}^{\infty} \frac{2^{n+1}}{\ln F_{n}} &= \frac{C}{\ln 2} \sum_{n=0}^{\infty} \frac{2^{n+1}}{\log_{2}(2^{2^{n}}+1)} \\ &> \frac{C}{\ln 2} \sum_{n=0}^{\infty} 1 \\ &= \infty,\end{align} \] and indeed this argument predicts that an asymptotically constant fraction of Fermat numbers are prime! There are a number of conditions that are equivalent to the primality of Fn. Proth's theorem (1878) — Let N = k2^{m} + 1 with odd k < 2^{m}. If there is an integer a such that \[ a^{(N1)/2} \equiv 1\mod N \! \] then N is prime. Conversely, if the above congruence does not hold, and in addition \[ \left(\frac{a}{N}\right)=1\! \](See Jacobi symbol) then N is composite. If N = F_{n} > 3, then the above Jacobi symbol is always equal to −1 for a = 3, and this special case of Proth's theorem is known as Pépin's test. Although Pépin's test and Proth's theorem have been implemented on computers to prove the compositeness of many Fermat numbers, neither test gives a specific nontrivial factor. In fact, no specific prime factors are known for n = 20 and 24. Let n ≥ 3 be a positive odd integer. Then n is a Fermat prime if and only if for every a coprime to n, a is a primitive root mod n if and only if a is a quadratic nonresidue mod n. \[ F_{n}=\left(2^{2^{n1}}\right)^{2}+1^{2}.\! \] When \[ F_{n} = x^2 + y^2 \] not of the form shown above, a proper factor is: \[ \gcd(x + 2^{2^{n1}} y, F_{n}).\! \] Example 1: F_{5} = 622642 + 204492, so a proper factor is \[ \gcd(62264\, +\, 2^{2^4}\cdot 20449,\, F_{5}) = 641.\! \] Example 2: F_{6} = 40468032562 + 14387937592, so a proper factor is \[ \gcd(4046803256\, +\, 2^{2^5}\cdot 1438793759,\, F_{6}) = 274177.\! \] Factorization of Fermat numbers Because of the size of Fermat numbers, it is difficult to factorize or to prove primality of those. Pépin's test gives a necessary and sufficient condition for primality of Fermat numbers, and can be implemented by modern computers. The elliptic curve method is a fast method for finding small prime divisors of numbers. Distributed computing project Fermatsearch has successfully found some factors of Fermat numbers. Yves Gallot's proth.exe has been used to find factors of large Fermat numbers. Édouard Lucas, improving the above mentioned result by Euler, proved in 1878 that every factor of Fermat number F_n, with n at least 2, is of the form k\times2^{n+2}+1 (see Proth number), where k is a positive integer; this is in itself almost sufficient to prove the primality of the known Fermat primes. Factorizations of the first twelve Fermat numbers are: 130,439,874,405,488,189,727,484,768,796,509,903,946,608,530,841,611,892,186,895,295,776,832,416,251,471,863,574, 173,462,447,179,147,555,430,258,970,864,309,778,377,421,844,723,664,084,649,347,019,061,363,579,192,879,108,857,591,038,330,408,837,177,983,810,868,451, As of February 2012, only F0 to F11 have been completely factored.[4] The distributed computing project Fermat Search is searching for new factors of Fermat numbers.[5] The set of all Fermat factors is A050922 (or, sorted, A023394) in OEIS. Like composite numbers of the form 2^{p} − 1, every composite Fermat number is a strong pseudoprime to base 2. Because all strong pseudoprimes to base 2 are also Fermat pseudoprimes  i.e. \[ 2^{F_n1} \equiv 1 \pmod{F_n}\,\! \] for all Fermat numbers. Because it is generally believed that all but the first few Fermat numbers are composite, this makes it possible to generate infinitely many strong pseudoprimes to base 2 from the Fermat numbers. In fact, Rotkiewicz showed in 1964 that the product of any number of prime or composite Fermat numbers will be a Fermat pseudoprime to base 2. Lemma: If n is a positive integer, \[ a^nb^n=(ab)\sum_{k=0}^{n1} a^kb^{n1k}. \] proof: \[ (ab)\sum_{k=0}^{n1}a^kb^{n1k} \] \[ =\sum_{k=0}^{n1}a^{k+1}b^{n1k}\sum_{k=0}^{n1}a^kb^{nk} \] \[ =a^n+\sum_{k=1}^{n1}a^kb^{nk}\sum_{k=1}^{n1}a^kb^{nk}b^n \] \[ =a^nb^n. \] Theorem: If \[ 2^n+1 \] is an odd prime, then n is a power of 2. proof: If n is a positive integer but not a power of 2, then n = rs where 1 \le r < n, 1 < s \le n and s is odd. By the preceding lemma, for positive integer m, \[ (ab) \mid (a^mb^m) \] where \mid means "evenly divides". Substituting a = 2^r, b = 1, and m = s and using that s is odd, \[ (2^r+1) \mid (2^{rs}+1), \] and thus \[ (2^r+1) \mid (2^n+1). \] Because \[ 1 < 2^r+1 < 2^n+1 \], it follows that \[ 2^n+1 \] is not prime. Therefore, by contraposition n must be a power of 2. Theorem: A Fermat prime cannot be a Wieferich prime. Proof: We show if \[ p=2^m+1 \] is a Fermat prime, then the congruence \[ 2^{p1} \equiv 1 \pmod {p^2} \] does not satisfy. It is easy to show 2m p1. Now write, \[ p1=2m\lambda\]. If the given congruence satisfies, then p^22^{2m\lambda}1, therefore \[ 0 \equiv (2^{2m\lambda}1)/(2^m+1)=(2^m1)(1+2^{2m}+2^{4m}+...+2^{2(\lambda1)m}) \equiv 2\lambda \pmod {2^m+1}.\ \] Hence \[ 2^m+12\lambda \] ,and therefore \[ 2\lambda \geq 2^m+1 \]. This leads to \[ p1 \geq m(2^m+1) \], which is impossible since \[ m \geq 2. \] A theorem of Édouard Lucas: Any prime divisor p of \[ F_n = 2^{2^{\overset{n}{}}}+1 \] is of the form k2^{n+2}+1 whenever n is greater than one. Sketch of proof: Let Gp denote the group of nonzero elements of the integers (mod p) under multiplication, which has order p1. Notice that 2 (strictly speaking, its image (mod p)) has multiplicative order \[ 2^{n+1} \] in Gp, so that, by Lagrange's theorem, p1 is divisible by \[ 2^{n+1} \] and p has the form k2^{n+1}+1 for some integer k, as Euler knew. Édouard Lucas went further. Since n is greater than 1, the prime p above is congruent to 1 (mod 8). Hence (as was known to Carl Friedrich Gauss), 2 is a quadratic residue (mod p), that is, there is integer a such that a2 2 is divisible by p. Then the image of a has order \[ 2^{n+2} \] in the group Gp and (using Lagrange's theorem again), p1 is divisible by \[ 2^{n+2} \] and p has the form \[ s2^{n+2}+1 \] for some integer s. In fact, it can be seen directly that 2 is a quadratic residue (mod p), since \[ (1 +2^{2^{n1}})^{2} \equiv 2^{1+2^{n1}} (mod p) \] . Since an odd power of 2 is a quadratic residue (mod p), so is 2 itself. An nsided regular polygon can be constructed with compass and straightedge if and only if n is the product of a power of 2 and distinct Fermat primes. In other words, if and only if n is of the form n = 2^{k}p_{1}p_{2}…p_{s}, where k is a nonnegative integer and the pi are distinct Fermat primes. A positive integer n is of the above form if and only if its totient φ(n) is a power of 2. Fermat primes are particularly useful in generating pseudorandom sequences of numbers in the range 1 … N, where N is a power of 2. The most common method used is to take any seed value between 1 and P − 1, where P is a Fermat prime. Now multiply this by a number A, which is greater than the square root of P and is a primitive root modulo P (i.e., it is not a quadratic residue). Then take the result modulo P. The result is the new value for the RNG. \[ V_{j+1} = \left( A \times V_j \right) \bmod P \](see Linear congruential generator, RANDU) This is useful in computer science since most data structures have members with 2X possible values. For example, a byte has 256 (28) possible values (0–255). Therefore to fill a byte or bytes with random values a random number generator which produces values 1–256 can be used, the byte taking the output value − 1. Very large Fermat primes are of particular interest in data encryption for this reason. This method produces only pseudorandom values as, after P − 1 repetitions, the sequence repeats. A poorly chosen multiplier can result in the sequence repeating sooner than P − 1. A Fermat number cannot be a perfect number or part of a pair of amicable numbers. (Luca 2000) The series of reciprocals of all prime divisors of Fermat numbers is convergent. (Křížek, Luca & Somer 2002) If n^{n} + 1 is prime, there exists an integer m such that \[ n = 2^{2^{m}} \]. The equation nn + 1 = F(2m+m) holds at that time.[6] Let the largest prime factor of Fermat number Fn be P(Fn). Then, \[ P(F_n)\ge 2^{n+2}(4n+9)+1. \] (Grytczuk, Luca & Wojtowicz 2001) Generalized Fermat numbers Numbers of the form \[ a^{2^{ \overset{n} {}}} + b^{2^{ \overset{n} {}}} \], where a > 1 are called generalized Fermat numbers. By analogy with the ordinary Fermat numbers, it is common to write generalized Fermat numbers of the form \[ a^{2^{ \overset{n} {}}} + 1 \] as F_{n}(a). In this notation, for instance, the number 100,000,001 would be written as F_{3}(10). An odd prime p is a generalized Fermat number if and only if p is congruent to 1 (mod 4) (with the exception of \[ 3 = 2^{2^{0}}+1) \]. Because of the ease of proving their primality, generalized Fermat primes have become in recent years a hot topic for research within the field of number theory. Many of the largest known primes today are generalized Fermat primes. Generalized Fermat numbers can be prime only for even a, because if a is odd then every generalized Fermat number will be divisible by 2. By analogy with the heuristic argument for the finite number of primes among the base2 Fermat numbers, it is to be expected that there will be only finitely many generalized Fermat primes for each even base. The smallest prime number Fn(a) with n > 4 is F5(30), or 3032+1. A more elaborate theory can be used to predict the number of bases for which Fn(a) will be prime for a fixed n. The number of generalized Fermat primes can be roughly expected to halve as n is increased by 1. Mersenne prime Notes ^ Křížek, Luca, Somer 2001, p. 38, Remark 4.15 References Golomb, S. W. (1963), "On the sum of the reciprocals of the Fermat numbers and related irrationalities", Canad. J. Math. 15: 475–478 Retrieved from "http://en.wikipedia.org/"

