費馬數
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其中為非負整數。
若是質數,可以得到n必須是2的冪。(若,其中且b為奇數,則,即是的因數。)也就是說,所有具有形式的質數必然是費馬數,這些質數稱為費馬質數。已知的費馬質數只有至五個。
基本性質
費馬數滿足以下的遞迴關係:
其中。這些等式都可以用數學歸納法推出。從最後一個等式中,我們可以推出哥德巴赫定理:任何兩個費馬數都沒有大於1的公因子。要推出這個,我們需要假設且和有一個公因子。那麼能把
和都整除;則能整除它們相減的差。因為,這使得。造成矛盾。因為所有的費馬數顯然是奇數。作為一個推論,我們得到質數個數無窮的又一個證明。
其他性質:
- 的位數可以表示成以為基數就是
- (參見高斯函數).
- 除了以外沒有費馬數可以表示成兩個質數的和。
- 當是奇質數的時候,沒有費馬數可以表示成兩個數的p次方相減的形式。
- 除了和,費馬數的最後一位是7。
- 所有費馬數(OEIS數列A051158)的倒數之和是無理數。 (所羅門·格倫布,1963)
費馬數的因式分解
最小的12個費馬數為:
F0 | = | 21 | + | 1 | = | 3 | |
F1 | = | 22 | + | 1 | = | 5 | |
F2 | = | 24 | + | 1 | = | 17 | |
F3 | = | 28 | + | 1 | = | 257 | |
F4 | = | 216 | + | 1 | = | 65,537 以上5個是已知的費馬質數。 | |
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, | ||||||
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 |
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= | 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, |
質因數個數顏色: 紅色:2個因數;綠色:3個因數;粉色:4個因數;藍色:5個因數;橘色:6個因數;紫色:7個因數(含以上);
只有最小的12個費馬數被人們完全分解了[1],目前最小不確定質性的費馬數是。
不確定質性的數:,n = 33、34、35、44、45、46、...
歷史
1640年,費馬提出了一個猜想,認為所有的費馬數都是質數。這一猜想對最小的5個費馬數成立,於是費馬宣稱他找到了表示質數的公式。然而,歐拉在1732年否定了這一猜想,他給出了的分解式:
歐拉證明費馬數的因數皆可表成,之後盧卡斯證明費馬數的因數皆可表成。
費馬的本猜測到了大數可說是大錯特錯,甚至多數數學家認為不存在第6個費馬質數。
定理
質性檢定
方法一
設為第個費馬數。如果不等於零,那麼:
- 是質數,若且唯若。
證明
假設以下等式成立:
那麼,因此滿足的最小整數一定整除,它是2的冪。另一方面,不能整除,因此它一定等於。特別地,存在至少個小於且與互質的數,這只能在是質數時才能發生。
假設是質數。根據歐拉準則,有:
- ,
其中是勒壤得符號。利用重複平方,我們可以發現,因此,以及。因為,根據二次互反律,我們便可以得出結論。
方法二
根據費馬平方和定理,可證明某些費馬數為合數。(因為型的質數皆可表示為兩正整數的平方和,且表示方法唯一) 例如:
注釋
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