|Jul18-12, 09:17 AM||#1|
Pells equation for D prime and =n*n-3
I, retired physicist (working with high level radioactive waste regulation) and now amateur mathematician, have been looking at solutions for the Pell equation
x*x-D*y*y=1, and I have in particular looked at the case D=n*n-3 which
contains solutions with high values for x and y, such as for D=61.
My simple studies have led me to formulate the following conjecture:
When D in Pell’s equation x*x-D*y*y=1 is
1) of the form n*n-3, and
2) a prime number
x+1 contains a set of factors which
a) includes D and
b) includes one or more factors from y twice, i.e. in a squared form.
For instance: x and y are the solution of x2-397y2=1 [x=838 721 786
045 180 184 649, y= 42 094 239 791 738 438 433 660] and
X+1 has the factors 2, 5^2, 17^2, 37^2, 173^2, 397, 1889^2, and
y the factors 2^2, 3^3, 5, 17, 37, 173, 383, 1 889, 990 151, of
which 5 appear as squares in x.
My idea is that if z=x+1 and (z-1)*(z-1)-Dy=1, then z*z-2z-Dy=0 and it
is possible to eliminate both D and more factors from the terms in the
equation, thereby revealing simpler relations between smaller terms.
(I also nourish a hope to find a general solution which will give me
the lowest solution, in additions to the other techniques that are
Is this something that you can prove false, or is it correct, perhaps
a well known fact?
Thank you for any comments you might have.
|Jul18-12, 09:38 AM||#2|
The example of x and y you gave doesn't work.
|Jul18-12, 03:49 PM||#3|
It works only with the floor function it would seem...
sqrt ((838721786045180184650^2 - 1)/397)
= 4.2094239791738433660 050188561322849559546514506... × 10^19
floor [sqrt ((838721786045180184650^2 - 1)/397)] = 42094239791738433660
UPDATE: With help from Wolfram Alpha...
x = ±1/2 (-(838721786045180184649-42094239791738433660 sqrt(397))^n-(838721786045180184649+42094239791738433660 sqrt(397))^n)
y = ±((838721786045180184649-42094239791738433660 sqrt(397))^n-(838721786045180184649+42094239791738433660 sqrt(397))^n)/(2 sqrt(397)), n element Z, n>=0
Closed form solutions for x and y, then, follow from simply substituting in integers for n.
x follows the form...
±((-(a - b*sqrt(397))^n - (a + b*sqrt(397))^n))/2
y follows the form...
±(((a - b*sqrt(397))^n - (a + b*sqrt(397))^n))/2*sqrt(397)
a = 838721786045180184649
b = 42094239791738433660
397, obviously, = 20^2 - 3 = n*n - 3
|Jul18-12, 04:56 PM||#4|
Pells equation for D prime and =n*n-3
I am not sure how to interprete the second comment, but perhaps I made an error with the very large numbers. I have not been able to copy and paste from Pari/gp.
Try instead n=8, n*n-3=61 x = 1 766 319 049 and y = 226 153 980
You will se the same pattern for x+1 (with 761 as the squared factor.)
Or you can try with D= 10*10-3 97 , x=62 809 633, y=6 377 352 (with 569)
or D=13 (with 5 a s the squared factor of x+1).
|Jul18-12, 05:24 PM||#5|
Let me just add that I got the D=397 solution (and all the others) from Wolfram alpha. I then inserted x+1 into Pari using the function factor(x+1) and then compared with factor(y).
|Jul18-12, 05:55 PM||#6|
Blog Entries: 2
|Jul18-12, 06:14 PM||#7|
(20-1)^2 + 4((20-2)/2) = 397
(18-1)^2 + 4((18-2)/2) = 321
Frame it that way and now you can insert into the Quadratic Formula:
(18-1)^2 - 4(-(18-2)/2) --> b^2 - 4ac = Discriminant
b = -(2ac + 1)
|Jul18-12, 06:48 PM||#8|
Also, Mickey, as a little mathematical factoid quite likely related to what you are doing... consider Lucas Sequences of U- and V- Type. One simple example is the Fibonacci Numbers (U) and the Lucas (V) Numbers. Another example is the Pell (U) Numbers and the Companion Pell (V) Numbers. A third example is the Mersenne (U) Numbers and the Fermat (V) Numbers.
Since U_n*V_n = U_2n, it then becomes clear that all factors of V_n are also factors of U_n, but not vice versa.
|Jul19-12, 10:18 AM||#9|
Thank you all for your help.
I don't think I can contribute to the statistics on high numbers to get data om high nn-3 primes. I suspect they get out of hand very soon. The results for D=397 were high enough. I might look a a few more using Wolfram alpha and Pari.
The reason I started to look at Pell's equation was that I looked for primes
p=4N+1=xx+yy, where xy divides 4N.
For these x and y we can write xx+yy=2nxy+1 since one of xor y must be even.
We can get a solution for x=ny+-sqrt((nn-1)y2+1) = ny+-z
if zz-(nn-1)yy=1, i.e. Pell's equation for D=nn-1.
This also connects to the Mersenne numbers 2^n-1 with a prime if for D=3
(Using brahmagupta's methodfor the two first solutions) I got some xx+yy=2nxy+1 which I thought more often rendered primes for D=3 than for the non-primes nn-1 (n>2), but my statistics is really poor, since I was using excel, I got 5 solutions or 3 Brahmagupta iterations.
I then saw that the third D below a square seems to give very high numbers.
I have another question: Since a prime of the type nn-3 = 4n+1 is composed of two squares, I looked at some cases like this: xx-(aa+bb)yy=1
It seems that a or b always (?) contain the factor 3.
Is that obvious to you?
|Jul20-12, 04:48 AM||#10|
I didn't have to work long on this problem to find that
xx-Dyy=1 implies that
xx-1=Dyy and therefore (x+1)(x-1)=Dyy
so that x+1 must contain at least some factor f of y and therefore also ff of yy.
What remains of my discovery is that D seems to be a factor of x+1 and not of x-1 at least for primes D=nn-3 (n=4, 8, 10,14, and 20) as far as I can see.
If true, it leads to another Pell-like equation with a lower solution.
|Oct3-12, 01:36 PM||#11|
Prof Hendrik from the Netherlands kindly provided me with the following proof for all d primes that x = -1 mod(d) (xx-dyy=1).
"I first note that x is odd, since if x is even then y is
odd, but now xx = 0 mod 4 and dyy + 1 = 2 mod 4, so that
xx and dyy + 1 cannot be equal.
So x is odd, and y must be even, say y = 2z. The two
positive integers w = (x - 1)/2 and w + 1 = (x + 1)/2 are
clearly coprime, and their product is d times a square. For
d prime, this can only happen if one of them is a square, say
uu, and the other is d times a square, say dvv. If w + 1
is the one that is a square, then we get uu = w + 1 = dvv + 1.
However, one has 0 < u < x, so this contradicts that x, y
is the LEAST positive solution to the Pell equation. Hence it
must be the other way around: w = uu, w + 1 = dvv, so that
x + 1 = 2(w + 1) is indeed divisible by d."
|Oct29-12, 05:16 PM||#12|
Let x(D) be a solution to Pell’s equation xx-Dyy=1 (x,y and D natural
it seems few people are concerned with x as a function of D. Here is a conjecture along the lines toward such an understanding, or perhaps rather pattern recognition.
If D is a prime = n(n+1)(n+2)+1 then
• x(D) is increasing with D
• x(D-1) is unusually low and x(D) is unusually high (NB the notorious case of D=61 is included), and
• n is approximately proportional the logarithm of x(D), see table
I realize there are a very limited number of cases presented, related to
my own limitation of tools. I simply counted the digits from Wolfram
Alpha's solutions. (Inclusion of non-primes would disturb the
Observe that the 3 degree of the polynomial for D rules out a simple
general polynomial representation of x, since the polynomial
representing xx would have an odd degree.
1st row: n / 2nd row: No of digits in x(D=n(n+1)(n+2)+1) / 3rd row: No of digits in x(D-1)
1 1 1
3 10 2
5 12 2
6 19 2
9 29 3
10 46 3
13 51 3
14 86 4
18 133 5
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