費馬數

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費馬數是以数学家费马命名一组自然数,具有形式:

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。(若n = ab,其中1 < a, b < nb为奇数,则2n + 1 ≡ (2a)b + 1 ≡ (−1)b + 1 ≡ 0(mod 2a + 1)。)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F0F4五個。

基本性质[编辑]

费马数满足以下的递回关系

F_{n} = (F_{n-1}-1)^{2}+1\,
F_{n} = F_{n-1} + 2^{2^{n-1}}F_{0} \cdots F_{n-2}
F_{n} = F_{n-1}^2 - 2(F_{n-2}-1)^2
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 < jFiFj 有一个公因子 a > 1。那么 a 能把

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

Fj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:

  • Fn的位数D(n,b)可以表示成以b基数就是
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。
  • 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中的数列A051158)的倒数之和是无理数。 (Solomon W. Golomb, 1963)

费马数的因式分解[编辑]

最小的12个費馬數为:

F0 = 21 + 1 = 3
F1 = 22 + 1 = 5
F2 = 24 + 1 = 17
F3 = 28 + 1 = 257
F4 = 216 + 1 = 65,537 以上5个是已知的费马素数。
F5 = 232 + 1 = 4,294,967,297
= 641 × 6,700,417
F6 = 264 + 1 = 18,446,744,073,709,551,617
= 274,177 × 67,280,421,310,721
F7 = 2128 + 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 = 2256 + 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 = 2512 + 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 = 21024 + 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 = 22048 + 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中的数列A000215)。

只有最小的12費馬數被完全分解[1]

历史[编辑]

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,他给出了分解式:

F5 = 232 + 1 = 4294967297 = 641 × 6700417

定理[编辑]

素性检验[编辑]

F_n=2^{2^n}+1为第n个费马数。如果n不等于零,那么:

F_n是素数,当且仅当3^{(F_n-1)/2}\equiv-1\pmod{F_n}

证明[编辑]

假设以下等式成立:

3^{(F_n-1)/2}\equiv-1\pmod{F_n}

那么3^{F_n-1}\equiv1\pmod{F_n},因此满足3k=1(modF_n)的最小整数k一定整除F_n-1=2^{2^n},它是2的幂。另一方面,k不能整除(F_n-1)/2,因此它一定等于F_n-1。特别地,存在至少F_n-1个小于F_n且与F_n互素的数,这只能在F_n是素数时才能发生。

假设F_n是素数。根据欧拉准则,有:

3^{(F_n-1)/2}\equiv\left(\frac3{F_n}\right)\pmod{F_n},

其中\left(\frac3{F_n}\right)勒让德符号。利用重复平方,我们可以发现2^{2^n}\equiv1\pmod3,因此F_n\equiv2\pmod3,以及\left(\frac{F_n}3\right)=-1。因为F_n\equiv1\pmod4,根据二次互反律,我们便可以得出结论\left(\frac3{F_n}\right)=-1

注释[编辑]

  1. ^ (英文) 費馬數的分解