h**********c
发帖数: 4120 | 1 尤其是第三部分，如果您能改进，希望能够合作发表。
认识有限，还望大牛不吝赐教。 | h**********c
发帖数: 4120 | | h**********c
发帖数: 4120 | | n***p
发帖数: 7668 | 4 愣是没看懂写的是啥。
【在 h**********c 的大作中提到】
: 搜索a^{\frac{1}{8}}好像也行。
| h**********c
发帖数: 4120 | 5 牛棚不好睡啊，胡说。
【在 n***p 的大作中提到】
: 愣是没看懂写的是啥。
| h**********c
发帖数: 4120 | 6 最后好像真的逼近2了。
【在 h**********c 的大作中提到】
: 搜索a^{\frac{1}{8}}好像也行。
| h**********c
发帖数: 4120 | 7 3. is not correct d1 d2 can > sqrt a
2. run 7 ~ 1000000 , 有两万个质数 sin(sqrt a pi ) <0，可能是运算精度的问题.有21万个合数（不含偶数）sin(sqrt a pi ) <0计算正确。 | h**********c
发帖数: 4120 | 8 3. 好像又对了
sin (\a \pi /x ) > 0
no roots or 4 roots, the least < a^{1/4}
If there are roots, then there must be d1 < d2 < a^{1/2}
case I. d2 has a factor of d1
then it can only be d1^2, otherwise there be another root before d2
so d1< a^{1/4}
case II. d2 doesn't have d1 as a factor, then d2 is prime.
d2 is prime then d1* d2 is also a factor. This d1* d2 won't pair up with the
two biggest root, since (d4= a/d1, d1*d2*d4 = a => d2 =1)
So except for the existing four roots, we have to add one more. If we add
one, we must add four. Observe the curve crossing x-axis. We now have eight
roots, four less than a^{1/2}, four greater.
So we have a third root d3 < a^{1/2}.
(1) If d3 has a factor of d1 or d2, it can only d1*d2, d1^2, or d1^2,
nothing else. So d1< a^{1/4}.
(2) If d3 does not has factor of d1 or d2, then d3 is prime. We know we also
has factor d4 < a^{1/2}.
d1, d2, d3 here are prime. d1*d2*d3 is a factor.(we have at least d1
here).
(2)-1 d4 has at least one of d1,d2,d3 as its factor
it can be a two from three combination or d1^2, d2^2 or d3^2 and no more,
think why? so d1< a^{1/4}.
(2)-2 d4 does divide any of d1,d2,d3,
d4* d3*d2* d1 <= a, in fact d4 is prime
d1< a^{1/4}.
QED?? | h**********c
发帖数: 4120 | 9 this can be the initialization b4 the loop.
the
【在 h**********c 的大作中提到】
: 3. 好像又对了
: sin (\a \pi /x ) > 0
: no roots or 4 roots, the least < a^{1/4}
: If there are roots, then there must be d1 < d2 < a^{1/2}
: case I. d2 has a factor of d1
: then it can only be d1^2, otherwise there be another root before d2
: so d1< a^{1/4}
: case II. d2 doesn't have d1 as a factor, then d2 is prime.
: d2 is prime then d1* d2 is also a factor. This d1* d2 won't pair up with the
: two biggest root, since (d4= a/d1, d1*d2*d4 = a => d2 =1)
| h**********c
发帖数: 4120 | 10 for the rsa, they use exactly two prime numbers, this is the case sin(\pi sqrt a) <0. We can use a binary search to locate sin(\pi a/x) == 0.
Somebody will not be easy. While the prime numbers can be bigger, even
binary search fatigue. | h**********c
发帖数: 4120 | 11 如果全部正确的话,够不够发一篇?
还是已经有这样的结论了? |
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