Like China, Egypt is also a famous ancient civilization in the world. When investigating the history of ancient Egypt, people noticed that a master of mathematics like Archimedes had actually studied Egypt's scores. Some of the greatest mathematicians of this century also studied Egyptian fractions. For example, wolf prize in mathematics laureate Paul Oudes put forward the famous conjecture 4/n =1/x+1/y+1/z, which stumped the world-class mathematicians. When nine loaves of bread were to be distributed equally to 10 people, the ancient Egyptians did not know that everyone could get 9/ 10, but said that everyone could get 1/3, 1/4, 1/5,1/. It is hard to imagine. You don't even know 9/ 10. How do you know 9/10 =1/3+1/4+1/5+112+65444? So for thousands of years, mathematical historians have insisted that ancient Egyptians did not use fractions.
1858, Scottish archaeologist Leiden bought an ancient Egyptian papyrus document. After identification, it is made of ponds formed by the flooding of the Nile and lush grass growing in swamps. The book was written about 1700 BC.
So, how did people in ancient Egypt calculate it? Divide the two items into four parts 1/2, give them to everyone first 1 1/2, and then divide the remaining 1 1/2 into three equal parts. Divide the results equally, and everyone will get 1/2 plus/kl. This "Leiden" papyrus paper, which has been preserved in the British Museum to this day, records the process of the decomposition of true fractions into monomolecular fractions in a large space. This operation method has been criticized by modern mathematicians, who believe that the complexity of fractional operation is one of the reasons why Egyptians failed to develop arithmetic and algebra to a higher level.
The pyramids in Egypt are world-famous, which shows that the ancient Egyptians had superb architectural skills and extraordinary wisdom. Don't you understand the simplest modern music score? Is the Golden Pagoda a shoddy work?
Modern mathematics has developed to a very abstract and complex level, but the scores of Egyptians are so rough that they should have disappeared from people's memory. However, the problems it caused still attract people's attention today.
Ke Zhao, the late old president of Sichuan University, once wrote: "Some problems caused by Egyptian symbols have become unsolved problems and conjectures, which have stumped many contemporary mathematicians." . Ke Zhao himself did not prove this conjecture until his death.
An old legend is:
On his deathbed, the old man gave 1 1 a horse to his three sons, the eldest 1/2, the second 1/4 and the third 1/6. Half of them are five and a half horses. We can't kill them. When we were helpless, our neighbors brought their own horses, and half of them brought six. The second quarter took three horses; One sixth of the third child brought two horses. A *** 1 1 horse, and the neighbor took his horse back after the division. That is1112 =1/2+1/4+1/6.
The wonderful Egyptian music finally mobilized its potential difficulty and defeated those who dared to despise them. And give embarrassing answers to those who laugh at him.
After more than two thousand years, mathematicians finally found: 2/n =1/[(n+1)/2]+1[(n+1) n/2]; 1/n = 1/(n+ 1)+ 1/[n(n+ 1)]； 1= 1/2+ 1/3+ 1/6。 At this time, I woke up from a big dream. Egyptian music stands in the world with great vitality, which amazed mathematicians after 3000 years. For example, can we design (n-1)/n =1/x+1/y+1/z ... After more than 2000 years' efforts, we finally uncovered the secret: there are six possibilities and seven ways of division. 7/8= 1/2+ 1/4+ 1/8; 1 1/ 12= 1/2+ 1/4+ 1/6= 1/2+ 1/3+ 1/ 12; 17/ 18= 1/2+ 1/3+ 1/9; 19/20= 1/2+ 1/4+ 1/5; 23/24= 1/2+ 1/3+ 1/8; 4 1/42= 1/2+ 1/3+ 1/7。 At first, people thought that there were probably infinite kinds of such situations, but it was really unpredictable to continue to pursue them but get nothing. 43 cases of * * were found in Guanchun River, Heilongjiang Province. This is correct.
When the denominator is odd, "1" is decomposed into Egyptian fraction, the number of terms is limited to 9, and * * * has five solutions:
1= 1/3+ 1/5+ 1/7+ 1/9+ 1/ 1 1+ 1/ 15+ 1/35+ 1/45+ 1/23 1。
1= 1/3+ 1/5+ 1/7+ 1/9+ 1/ 1 1+ 1/ 15+ 1/2 1+ 1/ 135+ 1/ 10395。
1= 1/3+ 1/5+ 1/7+ 1/9+ 1/ 1 1+ 1/ 15+ 1/2 1+ 1/ 165+ 1/693。
1= 1/3+ 1/5+ 1/7+ 1/9+ 1/ 1 1+ 1/ 15+ 1/2 1+ 1/23 1+ 1/3 15。
1= 1/3+ 1/5+ 1/7+ 1/9+ 1/ 1 1+ 1/ 15+ 1/33+ 1/45+ 1/385。
The above five groups of solutions are only found in 1976. When it is defined as 1 1, it is found that the smallest denominator of the solution of 1 is 105. If it is greater than 105, there are many solutions.
1/n type fraction can also be expressed as a series decomposition formula:
1/n= 1/(n+ 1)+ 1/(n+ 1)^2+ 1/(n+ 1)^3+ 1/(n+ 1)^4+....+ 1/(n+ 1)^k+ 1/n(n+ 1)^k.
The Egyptian fraction has become a dazzling pearl in the indefinite equation.
The most famous conjecture of Egyptian score is Erods conjecture: 1950. For positive integer n > 1, there is always 4/n =1/x+1/y+1/z (1).
Where x, y and z are all positive integers.
Stralss further conjectures that when n≥2, the solutions x, y and z of the equation satisfy x≠y, y≠z, z ≠ x.x < y < z. 。
In 1963, Zhao Ke, Sun Qi and Zhang Xianjue proved that the Erods conjecture is equivalent to the stralss conjecture. A few years later, yamanot developed the result to the seventh power of 10. Later, some mathematicians pushed the result forward and never got the fundamental solution. For 4/n = 1/x+ 1/z, we only need to consider the case that n=p is a prime number, because if (1) holds, for any integer m, m > 1,
4/pm =1/XM+1/ym+1/zm is also the same.
In 2002, someone put forward a stronger proposition: let x=AB, y=AC and Z=ABCp. (B)C)
4/P = 1/AB+ 1/AC+ 1/ABCP(2)
All odd prime numbers can be expressed as 4R+ 1 and 4R+3. For formula p=4R+3, formula (2) is obvious, because at this time A=(p+ 1)/4 and B= 1. C=P+ 1。 .
Namely: 4/p = {1[(p+1)/4]}+{1[(p+1) (p+1)/4]} {6544
For example: 4/7 =1/2+116+112.
For prime numbers of type p=4R+ 1, formula (2) is arranged as: 4ABC=PC+PB+ 1 (4).
A = (PC+PB+ 1)/4BC (5)
In formula (5), to get B|(PC+PB+ 1), it is necessary to make B|(PC+ 1) and let PC+1= TB; To get C|(PC+PB+ 1), let C|(PB+ 1) and Pb+1= sc; For the shape P=4R+ 1, if you want 4|p(C+B)+ 1], you need c+b = 4k-1; For the shape P=4R+3, 4|[P(C+B)+ 1] is required. Therefore, the binary linear indefinite equations are formed:
-PC+TB= 1 (6)
SC+(-P)B= 1 (7)
For example, when p= 17, A=3, B=2, C=5, T=43, S=7 and k=2.
4 / 17=[ 1/(2×3)]+[ 1/(3×5)]+[ 1/(3×2×5× 17 )]
That is, 4/17 =1/6+115+10.
Equivalent to the following formula:
(- 17)×5+43×2= 1
7×5+(- 17)×2= 1
Because for binary linear indefinite equations, we have a way. According to Shanghai Education Press 1985 (376 pages) Algebra Dictionary:
Equation: ax+by=c
a'x+b'y=c '
The necessary and sufficient conditions for the public * * * solution (integer solution) x and y are that (ab'-a'b) is not equal to 0, (ab'-a 'b) | (BC'-b 'c) | (ab'-a 'b )| (ca'-c 'a). "
Let's take C and B in (6) and (7) as X and Y in (6) above as long as (p, t) =1; B and c have infinite integer solutions; In (7), as long as (p, S)= 1, there are integer solutions of b and c. According to the known theorem (Ke Zhao, talking about indefinite equations) on pages 13 to 17, equations (6) and (7) must have a common integer. That is ST-P*P≠0, (ST-p * p) | (p+t); (ST-P*P) | (P+S). Why is there a solution? As long as one prime number has a solution, other prime numbers must have a solution. In China chess, the horse can jump to all points from the starting point, so the horse can jump to any point. Because horses can retreat from any point.
Here are some solutions to the p value:
- |
- p - | - A - | - B - | - C - | - T - | - S - | - K - |
- |
- 5 - | - 2 - | - 1 - | - 2 - | - 1 1 - | - 3 - | - 1 - |
-29 - | - 2 - | - 4 - | - 39 - | - 283 - | - 3 - | - 1 1 - |
-37 - | - 2 - | - 5 - | - 62 - | - 459 - | - 3 - | - 17 - |
-53 - | - 2 - | - 7 - | - 124 - | - 939 - | - 3 - | - 33 - |
-6 1 - | - 2 - | - 8 - | - 163 - | - 1243 - | - 3 - | - 43 - |
- 173-| - 2 - | - 22 - | - 1269 - | - 9979 - | - 3 - | - 323 - |
-
The above is the solution when P=4R+ 1 and r is an odd number. At this time, a = 2；; S=3 .
-
- 17 - | - 3 - | - 2 - | - 5 - | - 43 - | - 7 - | - 2 - |
-4 1 - | - 12 - | - 1 - | - 6 - | - 247 - | - 7 - | - 2 - |
-4 1 - | - 6 - | - 3 - | - 4 - | - 55 - | - 3 1 - | - 2 - |
-73 - | - 10 - | - 2 - | - 2 1 - | - 767 - | - 7 - | - 6 - |
- 97 - | - 17 - | - 2 - | - 5 - | - 243 - | - 39 - | - 2 - |
- 1 13-| - 5 - | - 6 - | - 97 - | - 1827 - | - 7 - | - 26 - |
-409-| - 59 - | - 2 - | - 13 - | - 2659 - | - 63 - | - 4 - |
-409-| - 22 - | - 5 - | - 66 - | - 5399 - | - 3 1 - | - 18 - |
-409-| - 1 1 - | - 1 1 - | - 60 - | - 223 1 - | - 75 - | - 18 - |
-
The above is the solution when p=4R+ 1 and r is an even number.
4 1 has two solutions; 409 has three solutions. That is to say, 4/41=1(12×1)+1(12× 6)+1(/kloc-0)
4/4 1= 1/(6×3)+ 1/(6×4)+ 1/(6×3×4×4 1)= 1/ 18+ 1/24+ 1/2952。
-4 1×6+247× 1= 1
7×6+(-4 1)× 1= 1
And a second set of solutions;
-4 1×4+55×3= 1
3 1×4+(-4 1×3)= 1
(2) The formula has solutions for all p values, but not all solutions. (For example, 4/4 1 has seven groups of solutions, while formula (2) only proves 4/p =1/ab+1/ac+1/abcp.
The formal solution of. Please pay attention to the difference between universal solution and all solutions.
In 1970s, people put forward the case of 5/P, and all prime numbers P can be expressed as 5R+ 1. 5R+2； 5R+3； 5R+4 shape.
For p = 5r+4,5/(5r+4) =1(r+1)+1[(5r+4) (r+1)].
Any one:1/n =1/(n+1)+1[n (n+1)].
For example, 5/9 =1/2+1/8 and1/2 =1/3+1/6; Or118 =119+1/(18×19).
For p = 5r+3,5/(5r+3) =1(r+1)+2/[(5r+3) (r+1)].
Any one of them: 2/n =1/[(n+1)/2]+1/[n (n+1)/2]
For example, 5/ 13= 1/3+2/39, 2/39 =1[(39+1)/2]+1[39× (39+60)
For p = 5r+2,5/(5r+2) =1(r+1)+3/[(5r+2) (r+1)].
R must be an odd number and (R+ 1) must be an even number.
and:3/[(5r+2)(r+ 1)]= 1/[(5r+2)(r+ 1)]+ 1/[(5r+2)(r+65438+)
For example, 5/37 =1/8+3/(37× 8); And 3/(37× 8) =1(37× 8)+1/(37× 4).
For P=5R+ 1,
Let 5/p =1/ab+1/AC+1/abcp (8).
5ABC=PC+PB+ 1 (9)
A=(PC+PB+ 1)/5BC ( 10)。
It can also be sorted into (6) and (7), and there are also solutions. B+C=5K- 1 shape.
The following are some solutions to the prime number of p=5R+ 1.
5/ 1 1 = 1/3+ 1/9+ 1/99，A=3，B= 1，C=3，T=34，S = 4；
5/3 1 = 1/7+ 1/56+ 1/ 1736，A=7，B= 1，C=8，T=248，S = 4；
5/4 1 = 1/9+ 1/93+ 1/ 1 1439，A=3，B=3，C=3 1，T=424，S = 4；
5/6 1 = 1/ 14+ 1/95+ 1/8 1 130，A= 1，B= 14，C=95，T=4 14，S = 9；
5/7 1 = 1/ 15+ 1/267+ 1/94785，A=3，B=5，C=89，T= 1264，S = 4；
5/ 10 1 = 1/2 1+ 1/53 1+ 1/3754 17，A=3，B=7，C= 177，T=2554，S = 4；
5/ 13 1 = 1/27+ 1/885+ 1/ 10434 15，A=3，B=9，C=295，T=4294，S = 4；
The method is the same as 4/P, please do it yourself.
Why must formulas (6) and (7) have solutions?
Two binary linear indefinite equations;
a 1x+b 1y= 1
a2x+b2y= 1。
The sufficient condition for a solution is (a1b2-a2b1) | (a1-a2); (a 1 B2-a2 b 1)|(B2-b 1)。
Let's examine a binary linear indefinite equation:
ax+by= 1。 ( 14)
According to the known theorem, as long as (a, b)= 1, (14) has the solution of the integer x, y, and there are infinite groups of solutions.
For example, 5x-2y= 1.
x； y
-
1, 2;
3, 7;
5, 12;
7, 17;
9, 22;
1 1,27;
13,32;
15,37;
17, 42;
19, 47;
...........
In other words, in the formula (14), x and y are also prime numbers. This is the basis for simultaneous equations to have a common solution. Now let's exchange x and y for a and b,
Take the above example as an example. Replace 5x-2y= 1 with 5a-2b= 1, x=5, y=2.
3x-7y= 1
17x-42y= 1
Form a binary binary linear indefinite equation.
5x- 12y= 1
19x-47y= 1
7x- 17y= 1
Form a ternary binary linear indefinite equation.
(4) The formula can be expressed as a prime formula:
P=(4ABC- 1)/(C+B). For example, when p=4 1, 41= (4x6x3x4-1)/(4+3); 4 1 =(5x3x 3x 3 1- 1)/(3 1+3)；
4 1 =(6x 1x8x 47- 1)/(8+47)； 4 1 =(7x 1x7x 36- 1)/(7+36)； 4 1 =(8x 6 x 1x 6- 1)/( 1+6)； 4 1 =(9x 1x6x 19- 1)/(6+ 19)；
4 1 =( 10x 1x6x 13- 1)/(6+ 13)； 4 1 =( 1 1x 1x4x 55- 1)/(4+55)； ; 4 1 =( 12x4x 1x 6- 1)/( 1+6)； ; 4 1 =( 13x 1x4x 15- 1)/(4+ 15)；
4 1 =( 14x 1x3x 124- 1)/(3+ 124)。 . It is valid until n =15: 41= (nabc-1)/(b+c).
People then asked: Is everything N?
p=(nABC- 1)/(B+C)。
Prime formula with three unknown variables can find all the prime numbers:
P=(4ABC- 1)/(B+C)。 ( 15)。
Equation (15) can be used for all prime numbers of the form p=4r+ 1
For example, 17. : 17 =(4x3x2x 5- 1)/(2+5)。
Equation (15) for all prime numbers of the form p=4r+3, A=(P+ 1)/4, B= 1, C=P+ 1 For example,11= (4x3x1x12-1)/(1+12).
For the form of complex number n=4r+3. n=(4xBXC- 1)/(B+C)。
For example, 51= (4x13x664-1)/(13+664). B=(P+ 1)/4，C=n(n+ 1)/4+ 1。
In fact, this problem is far from being solved.
Egyptian music score, an ancient discipline once despised by people, contains rich contents and many novel mysteries waiting for people to uncover.