費馬數

未解決的数学問題：当${\displaystyle n>4}$时，是否每个费马数都是合数？

費馬數是以数学家费马命名的一组自然数，具有形式：

${\displaystyle F_{n}=2^{2^{n}}+1}$

其中n为非负整数。

若2ⁿ + 1是素数，可以得到n必须是2的幂。（若n = ab，其中1 < a, b < n且b为奇数，则2ⁿ + 1 ≡ (2^a)^b + 1 ≡ (−1)^b + 1 ≡ 0(mod 2^a + 1)，即2^a + 1是2ⁿ + 1的因數。）也就是说，所有具有形式2ⁿ + 1的素数必然是費馬數，这些素数称为費馬素數。已知的費馬素數只有F₀至F₄五個。

基本性质

费马数满足以下的递归关系：

${\displaystyle F_{n}=(F_{n-1}-1)^{2}+1\,}$ 

${\displaystyle F_{n}=F_{n-1}+2^{2^{n-1}}F_{0}\cdots F_{n-2}}$ 

${\displaystyle F_{n}=F_{n-1}^{2}-2(F_{n-2}-1)^{2}}$ 

${\displaystyle F_{n}=F_{0}\cdots F_{n-1}+2}$ 

其中n ≥ 2。这些等式都可以用数学归纳法推出。从最后一个等式中，我们可以推出哥德巴赫定理：任何两个费马数都没有大于1的公因子。要推出这个，我们需要假设 0 ≤ i < j 且 F_i 和 F_j 有一个公因子 a > 1。那么 a 能把

${\displaystyle F_{0}\cdots F_{j-1}}$ 

和F_j都整除；则a能整除它们相减的差。因为a > 1，这使得a = 2。造成矛盾。因为所有的费马数显然是奇数。作为一个推论，我们得到素数个数无穷的又一个证明。

其他性质：

• F_n的位数D(n,b)可以表示成以b 为基数就是

${\displaystyle D(n,b)=\left\lfloor \log _{b}\left(2^{2^{\overset {n}{}}}+1\right)+1\right\rfloor \approx \lfloor 2^{n}\,\log _{b}2+1\rfloor }$  (参见高斯函数).

• 除了F₁ = 2 + 3以外没有费马数可以表示成两个素数的和。
• 当p是奇素数的时候，没有费马数可以表示成两个数的p次方相减的形式。
• 除了F₀和F₁，费马数的最后一位是7。
• 所有费马数（OEIS數列A051158）的倒数之和是无理数。 (Solomon W. Golomb, 1963)

费马数的因式分解

最小的12个費馬數为：

F0	=	2^1	+	1	=	3
F1	=	2^2	+	1	=	5
F2	=	2^4	+	1	=	17
F3	=	2^8	+	1	=	257
F4	=	2^16	+	1	=	65,537 以上5个是已知的费马素数。
F5	=	2^32	+	1	=	4,294,967,297
					=	641 × 6,700,417
F6	=	2^64	+	1	=	18,446,744,073,709,551,617
					=	274,177 × 67,280,421,310,721
F7	=	2^128	+	1	=	340,282,366,920,938,463,463,374,607,431,768,211,457
					=	59,649,589,127,497,217 × 5,704,689,200,685,129,054,721
F8	=	2^256	+	1	=	115,792,089,237,316,195,423,570,985,008,687,907,853,269,984,665,640,564,039,457,584,007,913,129,639,937
					=	1,238,926,361,552,897 × 93,461,639,715,357,977,769,163,558,199,606,896,584,051,237,541,638,188,580,280,321
F9	=	2^512	+	1	=	13,407,807,929,942,597,099,574,024,998,205,846,127,479,365,820,592,393,377,723,561,443,721,764,030,073,546, 976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,433,649,006,084,097
					=	2,424,833 × 7,455,602,825,647,884,208,337,395,736,200,454,918,783,366,342,657 × 741,640,062,627,530,801,524,787,141,901,937,474,059,940,781,097,519,023,905,821,316,144,415,759,504,705,008,092,818,711,693,940,737
F10	=	2^1024	+	1	=	179,769,313,486,231,590,772,930,519,078,902,473,361,797,697,894,230,657,273,430,081,157,732,675,805,500,963,132,708,477,322,407,536,021,120, 113,879,871,393,357,658,789,768,814,416,622,492,847,430,639,474,124,377,767,893,424,865,485,276,302,219,601,246,094,119,453,082,952,085, 005,768,838,150,682,342,462,881,473,913,110,540,827,237,163,350,510,684,586,298,239,947,245,938,479,716,304,835,356,329,624,224,137,217
					=	45,592,577 × 6,487,031,809 × 4,659,775,785,220,018,543,264,560,743,076,778,192,897 × 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, 140,227,977,573,104,895,898,783,928,842,923,844,831,149,032,913,798,729,088,601,617,946,094,119,449,010,595,906, 710,130,531,906,171,018,354,491,609,619,193,912,488,538,116,080,712,299,672,322,806,217,820,753,127,014,424,577
F11	=	2^2048	+	1	=	32,317,006,071,311,007,300,714,876,688,669,951,960,444,102,669,715,484,032,130,345,427,524,655,138,867,890,893,197,201,411,522,913,463,688,717, 960,921,898,019,494,119,559,150,490,921,095,088,152,386,448,283,120,630,877,367,300,996,091,750,197,750,389,652,106,796,057,638,384,067, 568,276,792,218,642,619,756,161,838,094,338,476,170,470,581,645,852,036,305,042,887,575,891,541,065,808,607,552,399,123,930,385,521,914, 333,389,668,342,420,684,974,786,564,569,494,856,176,035,326,322,058,077,805,659,331,026,192,708,460,314,150,258,592,864,177,116,725,943, 603,718,461,857,357,598,351,152,301,645,904,403,697,613,233,287,231,227,125,684,710,820,209,725,157,101,726,931,323,469,678,542,580,656, 697,935,045,997,268,352,998,638,215,525,166,389,437,335,543,602,135,433,229,604,645,318,478,604,952,148,193,555,853,611,059,596,230,657
					=	319,489 × 974,849 × 167,988,556,341,760,475,137 × 3,560,841,906,445,833,920,513 × 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, 546,421,940,712,978,306,134,189,864,280,826,014,542,758,708,589,243,873,685,563,973,118,948,869,399,158,545,506,611,147,420,216,132,557,017,260,564,139, 394,366,945,793,220,968,665,108,959,685,482,705,388,072,645,828,554,151,936,401,912,464,931,182,546,092,879,815,733,057,795,573,358,504,982,279,280,090, 942,872,567,591,518,912,118,622,751,714,319,229,788,100,979,251,036,035,496,917,279,912,663,527,358,783,236,647,193,154,777,091,427,745,377,038,294, 584,918,917,590,325,110,939,381,322,486,044,298,573,971,650,711,059,244,462,177,542,540,706,913,047,034,664,643,603,491,382,441,723,306,598,834,177

其中前八个来源于（OEIS數列A000215）。

只有最小的12个費馬數被人们完全分解了^[1]。

历史

1640年，费马提出了一个猜想，認為所有的费马数都是素数。这一猜想对最小的5个費馬數成立，于是费马宣称他找到了表示素数的公式。然而，欧拉在1732年否定了这一猜想，他给出了F₅的分解式：

F₅ = 2³² + 1 = 4294967297 = 641 × 6700417

歐拉證明費馬數的因數皆可表成k2ⁿ⁺¹ + 1，之後卢卡斯證明費馬數的因數皆可表成k2ⁿ⁺² + 1。

定理

主条目：可作图多边形

• 高斯称：尺规作图正多边形的边数目的充分条件是2的非負整數次方乘以任意个（可为0个）不同的费马素数的积，解决了兩千年来悬而未决的难题。這個條件也是必要條件，但他沒有給出證明。

素性检验

设${\displaystyle F_{n}=2^{2^{n}}+1}$ 为第n个费马数。如果n不等于零，那么：

${\displaystyle F_{n}}$ 是素数，当且仅当${\displaystyle 3^{\frac {F_{n}-1}{2}}\equiv -1{\pmod {F_{n}}}}$ 。

证明

假设以下等式成立：

${\displaystyle 3^{\frac {F_{n}-1}{2}}\equiv -1{\pmod {F_{n}}}}$ 

那么${\displaystyle 3^{F_{n}-1}\equiv 1{\pmod {F_{n}}}}$ ，因此满足3^k=1（mod${\displaystyle F_{n}}$ ）的最小整数k一定整除${\displaystyle F_{n}-1=2^{2^{n}}}$ ，它是2的幂。另一方面，k不能整除${\displaystyle {\tfrac {F_{n}-1}{2}}}$ ，因此它一定等于${\displaystyle F_{n}-1}$ 。特别地，存在至少${\displaystyle F_{n}-1}$ 个小于${\displaystyle F_{n}}$ 且与${\displaystyle F_{n}}$ 互素的数，这只能在${\displaystyle F_{n}}$ 是素数时才能发生。

假设${\displaystyle F_{n}}$ 是素数。根据欧拉准则，有：

${\displaystyle 3^{\frac {F_{n}-1}{2}}\equiv \left({\frac {3}{F_{n}}}\right){\pmod {F_{n}}}}$ ,

其中${\displaystyle \left({\frac {3}{F_{n}}}\right)}$ 是勒让德符号。利用重复平方，我们可以发现${\displaystyle 2^{2^{n}}\equiv 1{\pmod {3}}}$ ，因此${\displaystyle F_{n}\equiv 2{\pmod {3}}}$ ，以及${\displaystyle \left({\frac {F_{n}}{3}}\right)=-1}$ 。因为${\displaystyle F_{n}\equiv 1{\pmod {4}}}$ ，根据二次互反律，我们便可以得出结论${\displaystyle \left({\frac {3}{F_{n}}}\right)=-1}$ 。

注释

1. （英文） 費馬數的分解 （页面存档备份，存于互联网档案馆）