等差质数的猜想

[IT_worker]

猜想1：对于任何n总存在三个质数p1,p2,p3使得：p1>p2>p3>n，p1,p2,p3成等差数列
猜想2：对于任何n总存在四个质数p1,p2,p3,p4使得：p1>p2>p3>p4>n，p1,p2,p3,p3成等差数列
……………………………………

[medie2005]

不好意思，看错了，格林和陶哲轩的证明与楼主的命题不一致。

[db05s0511]

up
up

[medie2005]

哦，好象格林和陶哲轩的论文中也证明了由素数构成的长度为k的等差数列有任意多组：

“但是，他们得到的结果几乎是一个不能想象的伟大成就，他们证明由素数构成的等差数列可以任意长，而且有任意多组。4个数的素数等差数列可以有无穷多个的猜想都还没有证明，他们一下就跳这么远。”

以上均摘自《这项伟大的成就里有中国数学家的贡献--数学家王元谈菲尔茨奖获得者陶哲轩的工作》一文：
http://post.baidu.com/f?kz=151458812

[IT_worker]

to medie2005 : 格林和陶哲轩的结果显然蕴涵了我的猜想，真的不知道能够有这么棒的结论。
我贴这个猜想的来源出于我大二时候自己设计的一道题目：
猜想3：找出或者证明不存在四个平方数，它们成等差数列。
这个题目我一直都做不出来，甚至对于下面比它弱的多的题目
猜想4：总存在n，使得不可能存在n个成等差数列的平方数
由于质数的密度比平方数大的多，我想质数之间应该很容易找到等差数列了把，可是想了很久连猜想1也没有找到方法。
所以贴出了这篇帖子。
如果那位有兴趣，可以试着解决一下猜想3或者猜想4

[pptor]

关键在于构造出一样素数等差数列

[tailzhou]

没什么头绪；

平方数的最后位 只能是 0 1 4 5 6 9
能组成4个数的等差数列的个位只可能是
0 0 0 0
1 1 1 1
4 4 4 4
5 5 5 5
6 6 6 6
9 9 9 9
0 5 0 5
1 6 1 6
4 9 4 9
5 0 5 0
6 1 6 1
9 4 9 4
等差的差只能是5的倍数；

[C1053710211]

这里盗用一下尾巴的结论，在他的基础上化简了一下，
假设存在这样的四个平方数成等差数列，
于是我们设它们分别为x1,x2,x3,x4,它们是首项为a0^2，公差为d的等差数列，
这里b1是x2与x1底数的差，即x1=a0^2,x2=(a0+b1)^2，
所以,x2=a0^2+d=(a0+b1)^2=a0^2+b1^2+2*a0*b1 => d=b1^2++2*a0*b1;
同理可求 2d=b2^2+2*a0*b2, 3d=b3^2+2*a0*b3,
我们这里看2d=b2^2+2*a0*b2，等式左侧为偶数，右侧2*a0*b2为偶数，所以b2^2必须为偶数，
所以可以得出b2是偶数，又b2是偶数，所以b2^2是4的倍数，2*a0*b2也是4的倍数，所以d是2的倍数，
同理可以通过d=b1^2++2*a0*b1计算出b1是偶数，并且d必须是4的倍数。
上面的一大堆废话其实是要证明d是偶数而已，
至于是否还能推导出有用的结论我也没推下去，因为没有思路了，
d是偶数了，那么根据尾巴的结论d是5的倍数，我们有d是10的倍数，
也就是说，x1,x2,x3,x4的所有个位是相同的，现在就可以化简尾巴的结论为四个结尾数只能是
0，0，0，0，
1，1，1，1，
4，4，4，4，
5，5，5，5，
6，6，6，6，
9，9，9，9，
我们看第一列0，0，0，0，一定是由10的倍数产生的平方数，所以每个数都是100的倍数，
那么我们把它们都除以100后它们的商仍然是平方数，并且成等差数列，
所以第一种情况等价于其他情况解得并集，同理以4结尾的都是4，6结尾的都是4的倍数，
以5结尾的都是25的倍数，把他们不停的除以4或25，
直到他们的商的个位出现1，1，1，1或9，9，9，9的情况，就完成了化简，
所以源问题等价于是否存在尾数为如下两种情况的平方数成等差序列，
1，1，1，1
9，9，9，9
再往下就不会化简了，临帖涕零，不知所云
（说了一堆废话，其实也没证出来，看到这想骂我就骂我吧 ！！！）

[medie2005]

先贴一个三个完全平方数成等差数列时的结论。

三个完全平方数成等差数列等价于：丢番图方程x^2+y^2=2z^2的求解。
为简化，只考虑x,y,z为奇正整数且x<z<y的情形。

丢番图方程x^2+y^2=2z^2的所有整数解均可由：
x=m^2-n^2-2mn
y=m^2-n^2+2mn
z=m^2+n^2 （ m,n为整数,m>n,gcd(m,n)=1, m,n一奇一偶 ）
表示出。


设a^2， b^2， c^2， d^2为一等差数列。
并设：
c=(m^2-n^2+2mn)
a=(m^2-n^2-2mn)
b=(m^2+n^2) （ m,n为整数,m>n,gcd(m,n)=1, m,n一奇一偶 ）

设b=(p^2-q^2-2pq) ( p,q为整数, gcd(p,q)=1 p,q 一奇一偶 )
则:
c=p^2+q^2
d=(p^2-q^2+2pq)
d^2=(p^2-q^2+2pq)^2=p^4+4qp^3+2(pq)^2-4pq^3+q^4 (1)

因 b=(m^2+n^2)=(p^2-q^2-2pq)
将上式两边平方，得到：
m^4+2(nm)^2+n^4=p^4-4qp^3+2(pq)^2+4pq^3+q^4 (2)

而：d^2=2c^2-b^2=2(m^2-n^2+2mn)^2-(m^2+n^2)^2
=m^4+8nm^3+2(mn)^2-8mn^3+n^4 (3)

由 (1),(3)得：
p^4+4qp^3+2(pq)^2-4pq^3+q^4=m^4+8nm^3+2(mn)^2-8mn^3+n^4 (4)
而由(2)得到：
p^4-4qp^3+2(pq)^2+4pq^3+q^4=m^4+2(nm)^2+n^4 (5)

(4)-(5)得到：
8qp^3-8pq^3=8nm^3-8mn^3
即：
qp(p^2-q^2)=mn(m^2-n^2)
即：qp(p-q)(p+q)=mn(m-n)(m+n) (6)

因为gcd(p,q)=1, gcd(m,n)=1, 因此, q,p,(p-q),(p+q)两两互质； m,n,(m-n),(m+n)两两互质.

现在要证明(6)不成立，还没想到怎么证。

[Fogers]

看得头晕

[medie2005]

No Four Squares In Arithmetic ProgressionDickson's "History of the Theory of Numbers" presents a proof that
there do not exist four squares in arithmetic progression, but it
doesn't seem satisfactory. The problem with the proof (which Dickson
attributes to Bronwin and Furnass) is at this point:

... from the relations

y^2 - x^2 = z^2 - y^2 = w^2 - z^2

we know there are integers a,b,c,d such that

y+x = 2ab y-x = 2cd
z+y = 2ac z-y = 2bd
w+z = 2bc w-z = 2ad

Why must there exist four integers satisfying all six of these
relations? The rest of the proof relies entirely on this step, so
without some justification, the proof doesn't work. It isn't clear
(to me) how this approach can be fixed, so let's start at the
beginning and and try to give a valid elementary proof.

Suppose there exist four squares A^2, B^2, C^2, D^2 in increasing
arithmetic progression, i.e., we have B^2-A^2 = C^2-B^2 = D^2-C^2.
We can assume the squares are mutually co-prime, and the parity of
the equation shows that each square must be odd. Then we have co-prime
integers u,v such that A=u-v, C=u+v, u^2+v^2=B^2, and the common
difference of the progression is (C^2-A^2)/2=2uv.

We also have D^2 - B^2 = 4uv, which factors as [(D+B)/2][(D-B)/2] = uv.
The two factors on the left are co-prime, as are u and v, so there
exist four mutually co-prime integers a,b,c,d (exactly one even) such
that u=ab, v=cd, D+B=2ac, and D-B=2bd. This implies B=ac-bd, so we
can substitute into the equation u^2 + v^2 = B^2 to give
(ab)^2 + (cd)^2 = (ac-bd)^2.

This quadratic is symmetrical in the four variables, so we can assume
c is even and a,b,d are odd. From this quadratic equation we find that
c is a rational function of SQRT(a^4 - a^2 d^2 + d^4), which implies
there is an odd integer m such that a^4 - a^2 d^2 + d^4 = m^2.

Since a and d are odd there exist co-prime integers x and y such that
a^2 = k(x+y) and d^2 = k(x-y), where k=+-1. Substituting into the
above quartic gives x^2 + 3y^2 = m^2, from which it's clear that y must
be even and x odd. Changing the sign of x if necessary to make m+x
divisible by 3, we have 3(y/2)^2 = [(m+x)/2][(m-x)/2], which implies
that (m+x)/2 is three times a square, and (m-x)/2 is a square. Thus
we have co-prime integers r and s (one odd and one even) such that
(m+x)/2=3r^2, (m-x)/2=s^2, m=3r^2+s^2, x=3r^2-s^2, and y=+-2rs.

Substituting for x and y back into the expressions for a^2 and d^2 (and
transposing if necessary) gives a^2=k(s+r)(s-3r) and d^2=k(s-r)(s+3r).
Since the right hand factors are co-prime, it follows that the four
quantities (s-3r), (s-r), (s+r), (s+3r) must each have square absolute
values, with a common difference of 2r. These quantities must all have
the same sign, because otherwise the sum of two odd squares would equal
the difference of two odd squares, i.e., 1+1=1-1 (mod 4), which is
false.

Therefore, we must have |3r| < s, so from m=3r^2+s^2 we have 12r^2 < m.
Also the quartic equation implies m < a^2+d^2, so we have the inequality
|2r| < |SQRT(2/3)max(a,d)|. Thus we have four squares in arithmetic
progression with the common difference |2r| < |2abcd|, the latter being
the common difference of the original four squares. This contradicts
the fact that there must be a smallest absolute common difference for
four squares in arithmetic progression, so the proof is complete.

This theorem can be used to answer other questions as well. For
example, H. Jurjus asked whether there are rational numbers p, q such
that (p^2, q^2) is a point on the hyperbola given by (2-x)(2-y) = 1
with (p^2, q^2) not equal to (1,1). The answer is no. To see why,
suppose p=a/b and q=c/d (both fractions reduced to least terms).
Then if (p^2,q^2) is on the hyperbola we have

2b^2 - a^2 2d^2 - c^2
---------- ---------- = 1
b^2 d^2

Since b^2 is coprime to 2b^2-a^2 and c^2 is coprime to 2d^2-c^2 it
follows that b^2 = 2d^2 - c^2 and d^2 = 2b^2 - a^2. Rearranging
terms give

b^2 - d^2 = d^2 - c^2 d^2 - b^2 = b^2 - a^2

Together these equations imply that a^2, b^2, d^2 and c^2 are in
airthmetic progression, which is impossible.

[medie2005]

以上证明来自：http://www.mathpages.com/home/kmath044.htm

[C1053710211]

英文的好呀！！！looklook

[qrc1988]

牛人不止一两个呀 哈哈

[Vitin]

学习！

[rabbitbug]

来景仰牛人

[FancyMouse]

一问：质数密度到底是log(n)/n还是1/log(n)？偶记得结论是质数数目和n/log(n)同阶。。。

[xiaobukuai]

mark

[laomai]

顶！

[Vitin]

The Nth prime, P(N), is about N times the natural log of N:
P(N) ~ NlnN.

这样质数的密度应该是N/(NlnN) = 1/lnN.
平方数的密度则为sqrt(N)/N = 1/sqrt(N).
所以确实是质数的密度大．

[deng2000]

惭愧之至,我记错了素数定理的结果. FancyMouse和Vitin的理解是正确的,感谢二位指正.