回答|共 19 个

仙都峰居士 LV6

发表于 19-6-2012 18:38:55 | 显示全部楼层

来自中一书本上的题目:
"Every even number greater than two can be expressed as a sum of two prime numbers". Do you agree? Why?

davidbin LV6

发表于 20-6-2012 16:07:26 | 显示全部楼层

小狮租房
Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:
Every even integer greater than 2 can be expressed as the sum of two primes.

jjrchome LV16

发表于 20-6-2012 17:17:30 | 显示全部楼层

仙都峰居士 发表于 19-6-2012 18:38
来自中一书本上的题目:
"Every even number greater than two can be expressed as a sum of two prime nu ...

这个就是强哥德巴赫猜想 ---“ 数学皇冠上的明珠” ,也就是我们常说的“1+1”。这个猜想至今无人能够证明。

当年陈景润用了六麻袋的草稿纸才证明出了的“1+2”(任何一个足够大的偶数都可以表示成一个素数和一个半素数的和),随后穷尽一生也无法解出“1+1”。

“强哥德巴赫猜想”或许永远也无法成为定理了。

仙都峰居士 LV6

发表于 21-6-2012 08:36:41 | 显示全部楼层

谢谢两位的解答,很想知道六麻袋的草稿纸里写的是什么。

仙都峰居士 LV6

发表于 24-6-2013 18:17:46 | 显示全部楼层

两个问题请大侠帮忙, 谢谢!
1. Prove that there are no integers a, b and c that satisfy the equation a^3+b^3=7c^3+3

2. Two circles intersect at point A and B, a common tangent touches the first circle at point C and the second at point D. Let B be inside the triangle ACD. Let the line CB intersect the second circle again at point E.
    Prove that AD bisects the angle CAE.

frekiwang LV15

发表于 24-6-2013 21:28:09 | 显示全部楼层

1.
如果x被7除余0,1,2,3,4,5,6; 那么x^3被7除分别余0,1,1,6,1,6,6

左边a^3+b^3被7除余数只可能为0+0 (0),0+1 (1),1+1 (2), 0+6 (6), 1+6 (0), 6+6 (5)
右边7c^3+3,被7除余数只可能为3.

frekiwang LV15

发表于 24-6-2013 21:28:45 | 显示全部楼层

仙都峰居士 发表于 24-6-2013 18:17
两个问题请大侠帮忙, 谢谢!
1. Prove that there are no integers a, b and c that satisfy the equation a ...

1.
如果x被7除余0,1,2,3,4,5,6; 那么x^3被7除分别余0,1,1,6,1,6,6

左边a^3+b^3被7除余数只可能为0+0 (0),0+1 (1),1+1 (2), 0+6 (6), 1+6 (0), 6+6 (5)
右边7c^3+3,被7除余数只可能为3.

frekiwang LV15

发表于 24-6-2013 21:40:02 | 显示全部楼层

仙都峰居士 发表于 24-6-2013 18:17
两个问题请大侠帮忙, 谢谢!
1. Prove that there are no integers a, b and c that satisfy the equation a ...

2.
Q2.jpg

静雯 LV7

发表于 24-6-2013 22:39:08 | 显示全部楼层

本帖最后由 静雯 于 24-6-2013 23:04 编辑

Refer to Frekiwang's picture.

Join AB and BD, we can get angle BCD =  angle BAC and angle BDC =  angle BAD  as CD is a common tangent to the two circles.

Angle CAD = angle BAC + angle BAD,  angle DBE = angle BCD + angle BDC.
So angle CAD = angle DBE.

Angle DBE = angle DAE as they share the same arc DE.

Thus angle CAD = angle DAE.

Proved.

(Thanks to Frekiwang for his picture.)


仙都峰居士 LV6

发表于 26-6-2013 17:28:57 | 显示全部楼层

多谢两位解答, 感激不尽!
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