Editing Fermat number (section)

==Primality==
Fermat numbers and Fermat primes were first studied by Pierre de Fermat, who [[conjecture]]d that all Fermat numbers are prime. Indeed, the first five Fermat numbers ''F''<sub>0</sub>, ..., ''F''<sub>4</sub> are easily shown to be prime. Fermat's conjecture was refuted by [[Leonhard Euler]] in 1732 when he showed that

:<math> F_{5} = 2^{2^5} + 1 = 2^{32} + 1 = 4294967297 = 641 \times 6700417. </math>

Euler proved that every factor of ''F''<sub>''n''</sub> must have the form {{nowrap|''k''{{space|hair}}2<sup>''n''+1</sup> + 1}} (later improved to {{nowrap|''k''{{space|hair}}2<sup>''n''+2</sup> + 1}} by [[Édouard Lucas|Lucas]]) for {{nowrap|''n'' ≥ 2}}.

That 641 is a factor of ''F''<sub>5</sub> can be deduced from the equalities 641&nbsp;=&nbsp;2<sup>7</sup>&nbsp;×&nbsp;5&nbsp;+&nbsp;1 and 641&nbsp;=&nbsp;2<sup>4</sup>&nbsp;+&nbsp;5<sup>4</sup>. It follows from the first equality that 2<sup>7</sup>&nbsp;×&nbsp;5&nbsp;≡&nbsp;−1&nbsp;(mod&nbsp;641) and therefore (raising to the fourth power) that 2<sup>28</sup>&nbsp;×&nbsp;5<sup>4</sup>&nbsp;≡&nbsp;1&nbsp;(mod&nbsp;641). On the other hand, the second equality implies that 5<sup>4</sup>&nbsp;≡&nbsp;−2<sup>4</sup>&nbsp;(mod&nbsp;641). These [[Modular arithmetic|congruences]] imply that 2<sup>32</sup>&nbsp;≡&nbsp;−1&nbsp;(mod&nbsp;641).

Fermat was probably aware of the form of the factors later proved by Euler, so it seems curious that he failed to follow through on the straightforward calculation to find the factor.<ref>{{Harvnb|Křížek|Luca|Somer|2001|p=38, Remark 4.15}}</ref> One common explanation is that Fermat made a computational mistake.

There are no other known Fermat primes ''F''<sub>''n''</sub> with {{nowrap|''n'' > 4}}, but little is known about Fermat numbers for large ''n''.<ref>Chris Caldwell, [http://primes.utm.edu/links/theory/special_forms/ "Prime Links++: special forms"] {{webarchive|url=https://web.archive.org/web/20131224224552/http://primes.utm.edu/links/theory/special_forms/ |date=2013-12-24 }} at The [[Prime Pages]].</ref> In fact, each of the following is an open problem:
* Is ''F''<sub>''n''</sub> [[composite number|composite]] [[for all]] {{nowrap|''n'' > 4}}?
* Are there infinitely many Fermat primes? ([[Gotthold Eisenstein|Eisenstein]] 1844<ref>{{Harvnb|Ribenboim|1996|p=88}}.</ref>)
* Are there infinitely many composite Fermat numbers?
* Does a Fermat number exist that is not [[square-free number|square-free]]?

{{As of|2024}}, it is known that ''F''<sub>''n''</sub> is composite for {{nowrap|5 ≤ ''n'' ≤ 32}}, although of these, complete factorizations of ''F''<sub>''n''</sub> are known only for {{nowrap|0 ≤ ''n'' ≤ 11}}, and there are no known prime factors for {{nowrap|1=''n'' = 20}} and {{nowrap|1=''n'' = 24}}.<ref name="Keller"/> The largest Fermat number known to be composite is ''F''<sub>18233954</sub>, and its prime factor {{nowrap|7 × 2<sup>18233956</sup> + 1}} was discovered in October 2020.

===Heuristic arguments===
Heuristics suggest that ''F''<sub>4</sub> is the last Fermat prime.

The [[prime number theorem]] implies that a random integer in a suitable interval around ''N'' is prime with probability 1{{space|hair}}/{{space|hair}}ln ''N''. If one uses the heuristic that a Fermat number is prime with the same probability as a random integer of its size, and that ''F''<sub>5</sub>, ..., ''F''<sub>32</sub> are composite, then the expected number of Fermat primes beyond ''F''<sub>4</sub> (or equivalently, beyond ''F''<sub>32</sub>) should be 
:<math> \sum_{n \ge 33} \frac{1}{\ln F_{n}} < \frac{1}{\ln 2} \sum_{n \ge 33} \frac{1}{\log_2(2^{2^n})} = \frac{1}{\ln 2} 2^{-32} < 3.36 \times 10^{-10}.</math>
One may interpret this number as an upper bound for the probability that a Fermat prime beyond ''F''<sub>4</sub> exists.

This argument is not a rigorous proof. For one thing, it assumes that Fermat numbers behave "randomly", but the factors of Fermat numbers have special properties. Boklan and [[John H. Conway|Conway]] published a more precise analysis suggesting that the probability that there is another Fermat prime is less than one in a billion.<ref>{{Cite journal |last1=Boklan |first1=Kent D. |last2=Conway |first2=John H. |date=2017 |title=Expect at most one billionth of a new Fermat Prime! |journal=The Mathematical Intelligencer |volume=39 |issue=1 |pages=3–5 |arxiv=1605.01371 |doi=10.1007/s00283-016-9644-3|s2cid=119165671 }}</ref>

Anders Bjorn and [[Hans Riesel]] estimated the number of square factors of Fermat numbers from ''F''<sub>5</sub> onward as
:<math>
\sum_{n \ge 5} \sum_{k \ge 1} \frac{1}{k (k 2^n + 1) \ln(k 2^n)} < \frac{\pi^2}{6 \ln 2} \sum_{n \ge 5} \frac{1}{n 2^n} \approx 0.02576;
</math>
in other words, there are unlikely to be any non-squarefree Fermat numbers, and in general square factors of <math>a^{2^n} + b^{2^n}</math> are very rare for large ''n''.<ref name="bjorn">{{cite journal |last1=Björn |first1=Anders |last2=Riesel |first2=Hans |title=Factors of generalized Fermat numbers |journal=Mathematics of Computation |date=1998 |volume=67 |issue=221 |pages=441–446 |doi=10.1090/S0025-5718-98-00891-6 |url=https://www.ams.org/journals/mcom/1998-67-221/S0025-5718-98-00891-6/ |language=en |issn=0025-5718|doi-access=free }}</ref>

===Equivalent conditions===
{{Main|Pépin's test}}

Let <math>F_n=2^{2^n}+1</math> be the ''n''th Fermat number. Pépin's test states that for {{nowrap|''n'' > 0}},

:<math>F_n</math> is prime if and only if <math>3^{(F_n-1)/2}\equiv-1\pmod{F_n}.</math>

The expression <math>3^{(F_n-1)/2}</math> can be evaluated modulo <math>F_n</math> by [[exponentiation by squaring|repeated squaring]]. This makes the test a fast [[polynomial-time]] algorithm. But Fermat numbers grow so rapidly that only a handful of them can be tested in a reasonable amount of time and space.

There are some tests for numbers of the form {{nowrap|''k''{{space|hair}}2<sup>''m''</sup> + 1}}, such as factors of Fermat numbers, for primality.

:'''[[Proth's theorem]]''' (1878). Let {{nowrap|1=''N'' = ''k''{{space|hair}}2<sup>''m''</sup> + 1}} with odd {{nowrap|''k'' < 2<sup>''m''</sup>}}. If there is an integer ''a'' such that
:: <math>a^{(N-1)/2} \equiv -1\pmod{N}</math>
:then <math>N</math> is prime. Conversely, if the above congruence does not hold, and in addition
:: <math>\left(\frac{a}{N}\right)=-1</math> (See [[Jacobi symbol]])
:then <math>N</math> is composite.

If {{nowrap|1=''N'' = ''F''<sub>''n''</sub> > 3}}, then the above Jacobi symbol is always equal to −1 for {{nowrap|1=''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 some Fermat numbers, neither test gives a specific nontrivial factor. In fact, no specific prime factors are known for {{nowrap|1=''n'' = 20}} and&nbsp;24.