- 余辉
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Sierpiński 的初等数论问题
波兰数学家 Wacu0142aw Sierpiński 对数论有很多研究。在他一生出版的 50 多本书里, 250 Problems of Elementary Number Theory 一书显得格外有趣。这里面不但有各种出人意料的数学事实,还有很多精妙的证明和大胆的构造,让人大呼过瘾。我从中选择了一些问题,在这里和大家一块儿分享。下面的文字没有完全照搬书中的内容,而是做了大量的改动和扩展;若有出错的地方,还请大家指正。个别题目会涉及一些初等数论中的著名定理,它们都可以在这篇文章里找到。
找出所有的正整数 n ,使得 n2 + 1 能被 n + 1 整除。
满足要求的解只有一个: n = 1 。原因很简单:如果 n2 + 1 = n(n + 1) – (n – 1) 是 n + 1 的整倍数,那么 n – 1 也必须是 n + 1 的整倍数,这只有一种可能性,即 n – 1 = 0 。
证明:对于任意大于 6 的偶数 n ,我们都能找到两个质数 p 和 q ,使得 n – p 和 n – q 互质。
不管 n 是多少,令 p = 3, q = 5 即可。这样一来, n – p 和 n – q 就是两个相邻的奇数,它们必然互质。
找出所有公差为 100 的等差数列,使得里面的所有项都是质数。
满足要求的等差数列不存在。这是因为,在 p, p + 100, p + 200 这三个数当中,至少有一个数能被 3 整除,因而 p 只能等于 3 。此时, p + 200 = 3 + 200 = 203 = 7 × 29 ,这就说明满足要求的等差数列不存在。
找出所有这样的质数,它既能表示成两个质数之和,也能表示成两个质数之差。
满足要求的数只有 5 ,它可以表示成 3 + 2 和 7 – 2 。下面我们证明,这个问题没有别的解了。如果质数 r 能表示成两个质数之和,那么显然 r > 2 ,因而 r 只能是奇数。两个质数之和是一个奇数,则其中一个质数一定是 2 ;两个质数之差是一个奇数,则其中一个质数也一定是 2 。因此, r 只有可能被表示成 p + 2 和 q – 2 ,其中 p 和 q 都是质数。这说明, p, r, q 是三个连续奇数。三个连续奇数当中,必然有一个能被 3 整除。如果它们都是质数,那么一定有一个数就是 3 。因此, (p, r, q) = (3, 5, 7) 是唯一的可能。
33 = 3 × 11 , 34 = 2 × 17 , 35 = 5 × 7 。它们组成了三个连续的正整数,其中每个数都是两个不同的质数之积。是否存在四个连续的正整数,使得每个数都是两个不同的质数之积?
不存在。任意四个连续的正整数中,一定有一个能被 4 整除,它显然不是两个不同的质数之积。
证明:方程 xy + x + y = 232 存在正整数解。
原方程相当于 xy + x + y + 1 = 232 + 1 ,即 (x + 1) · (y + 1) = 225 + 1 ,而后者是 n = 5 时的 Fermat 数,众所周知,它是能被分解成两个大于 1 的整数之积的。
证明:方程 x2 + y2 + 1 = z2 有无穷多组正整数解。
对于任意正整数 n , (2n)2 + (2n2)2 + 1 = (2n2 + 1)2 都成立。
证明:对于任意一个无限小数(不一定是无限循环小数),我们都能找到一个任意长的数字串,使得它会在这个无限小数的小数展开当中出现无穷多次。
令 m 为任意大的正整数。把小数点后的数字每 m 位分成一组,从而得到无穷多个 m 位数字串。由于不同的 m 位数字串只有 10m 种,因而必然有一种数字串会出现无穷多次。
证明:对于任意正整数 m ,总存在一个关于 x 和 y 的整系数方程 ax + by = c ,使得方程恰好有 m 个正整数解。
不管 m 是多少,令 c = m + 1 ,则方程 x + y = c 满足要求。这个方程显然有且仅有 m 个解,它们分别是 (1, m), (2, m – 1), …, (m, 1) 。
- 里论外几
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A search was also carried out for 5th power solutions. For example:
Are there any solutions to I^5 + J^5 = K^5 + L^5 ?
A search was made out to 3.6E+28 for fifth power solutions. No solutions were found. Again it might be that the computer program was faulty, or solutions might exist above 3.6E+28. The computer program that was used was essentially the same as the posted ramanujans.c program that was referenced earlier - except the cubes tables were replaced by fifth powers. For the fifth power search, regular “double” variables were replaced by “long double”. The extra precision allowed the search to run to 3.6E+28.
A similar search was run for sixth power solutions out to 3.6E+28. (e.g. I^6 + J^6 = K^6 + L^6) Nothing was found here either.1) 133^4 + 134^4 = 59^4 + 158^4 = 635,318,657 (Primitive)
2) 157^4 + 227^4 = 7^4 + 239^4 = 3,262,811,042 (Primitive)
3) 256^4 + 257^4 = 193^4 + 292^4 = 8,657,437,697 (Primitive)
4) 266^4 + 268^4 = 118^4 + 316^4 = 10,165,098,512 (A multiple of # 1)
5) 399^4 + 402^4 = 177^4 + 474^4 = 51,460,811,217 (A multiple of # 1)
Taxicab(6) = 24153319581254312065344
= 28906206^3 + 582162^3
= 28894803^3 + 3064173^3
= 28657487^3 + 8519281^3
= 27093208^3 + 16218068^3
= 26590452^3 + 17492496^3
= 26224366^3 + 18289922^3
1,847,282,122^ 3 + 2,648,660,966^3 =
1,766,742,096^3 + 2,685,635,652^3 =
1,638,024,868^3 + 2,736,414,008^3 =
860,447,381^3 + 2,894,406,187^3 =
459,531,128^3 + 2,915,734,948^3 =
309,481,473^3 + 2,918,375,103^3 =
58,798,362^3 + 2,919,526,806^3 =
24,885,189,317,885,898,975,235,988,544
Cabtaxi(1) = 1
= 1^3 + 0^3
Cabtaxi(2) = 91
= 3^3 + 4^3
= 6^3 - 5^3
Cabtaxi(3) = 728
= 6^3 + 8^3
= 9^3 - 1^3
= 12^3 - 10^3
Cabtaxi(4) = 2,741,256
= 108^3 + 114^3
= 140^3 - 14^3
= 168^3 - 126^3
= 207^3 - 183^3
Cabtaxi(5) = 6,017,193
= 166^3 + 113^3
= 180^3 + 57^3
= 185^3 - 68^3
= 209^3 - 146^3
= 246^3 - 207^3
Cabtaxi(6) = 1,412,774,811
= 963^3 + 804^3
= 1,134^3 - 357^3
= 1,155^3 - 504^3
= 1,246^3 - 805^3
= 2,115^3 - 2,004^3
= 4,746^3 - 4,725^3
Cabtaxi(7) = 11,302,198,488
= 1,926^3 + 1,608^3
= 1,939^3 + 1,589^3
= 2,268^3 - 714^3
= 2,310^3 - 1,008^3
= 2,492^3 - 1,610^3
= 4,230^3 - 4,008^3
= 9,492^3 - 9,450^3
Cabtaxi(8) = 137,513,849,003,496
= 22,944^3 + 50,058^3
= 36,547^3 + 44,597^3
= 36,984^3 + 44,298^3
= 52,164^3 - 16,422^3
= 53,130^3 - 23,184^3
= 57,316^3 - 37,030^3
= 97,290^3 - 92,184^3
= 218,316^3 - 217,350^3
Cabtaxi(9) = 424,910,390,480,793,000
= 645,210^3 + 538,680^3
= 649,565^3 + 532,315^3
= 752,409^3 - 101,409^3
= 759,780^3 - 239,190^3
= 773,850^3 - 337,680^3
= 834,820^3 - 539,350^3
= 1,417,050^3 - 1,342,680^3
= 3,179,820^3 - 3,165,750^3
= 5,960,010^3 - 5,956,020^3
Cabtaxi(10) has been confirmed by the author"s computer program and is equal to:
933,528,127,886,302,221,000
= 7,002,840^3 + 8,387,730^3
= 6,920,095^3 + 8,444,345^3
= 77,480,130^3 - 77,428,260^3
= 41,337,660^3 - 41,154,750^3
= 18,421,650^3 - 17,454,840^3
= 10,852,660^3 - 7,011,550^3
= 10,060,050^3 - 4,389,840^3
= 9,877,140^3 - 3,109,470^3
= 9,781,317^3 - 1,318,317^3
= 9,773,330^3 - 84,560^3
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