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Topic: Mersenne, Pell and Ramanujan end more (Read 3354 times) |
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Mickey1
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Mersenne, Pell and Ramanujan end more
« on: Mar 3rd, 2013, 7:28am » |
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Pell’s equation (PE), xx-dyy=1, has the highest solutions for x and y among d<100 for d=61. Looking at various ways to incorporate d=61 in a group of other values of d includes the approaches d=nn-3, and d=n(n+1)(n+2)+1. An observation is that x for primes d=n(n+1)(n+2) is usually much lower than for d=n(n+1)(n+2)+1. (Prof John Robertson has told me – without demonstration - in an email that counter-examples exist. I assume they are very high, certainly higher than d= 1560781 = 115*116*117+1. In this region the x solutions for d=n(n+1)(n+2)+1are often in excess of 1E+300, and very much lower for d=n(n+1)(n+2).
A third approach, presented here (and perhaps already known to you), is made by looking at the equation for d=2^n-3, hence the - not quite perfect – connection to the Mersenne numbers, including d= 5, 13, 29, 61, 125, 253..
The sometime large solutions can be generated by solving the companion equation with lower solutions
ss-dtt=-1 (eq. 1)
xx-dyy=1 for [x,y] =[2ss+1, 2st]
Another equation with even lower solutions is
uu-dvv=-4 (eq. 2)
xx-dyy=1 for [x,y] =[ (u2 + 3)u/2, (u2 + 1)v/2] (u,v,odd)
Looking at solutions for equation 1, I noted that for some n, t in equation 1 was simply equal to the previous d in the sequence d=2^n-3 i.e. the Mersenne numbers minus 2, i.e. we have for
N d=2^n-3 d=2^(n-1)-3 Pell's equation
___________________/_______________________/
3_________5_______1_________________2^2-5*1^2=-1
4_________13_____ 5_______________18^2-13*5^2 =-1
5_________29_____13______________70^2-29*13^2=-1
6_________61_____29_______________no solution
7________125_____61___________682^2-125^*61^2=-1
15_____32765_ _16381___2965142^2-32765*16381^2=-1
The reader can appreciate my disappointment that 61 came up only in the solution for n=7 with n=6 without solution.
The general formulation of the equations is ss-dtt = ss - (2^n-3)*(2^(n-1)-3)^2=-1 requiring that ss= (2^n-3)*(2^(n-1)-3)^2-1 = 0, and therefore that
(2^n-3)*(2^(n-1)-3)^2-1 must be a square of a natural number.
After some rearrangement (and credit to Wolfram Alpha) we have
(2^n-3)*(2^(n-1)-3)^2-1 = 1/4 (-7+2^n) (-4+2^n)^2, i.e. it is required that
2^n-7 is a perfect square, 2^n-7=x^2. This is the Ramanujan-Nagell equation, conjectured by Ramanujan (that solutions exist for 3, 4, 5, 7 and 15 only), proposed independently in 1943 by Wilhelm Ljunggren, and proved in 1948 by Trygve Nagell, (Wikipedia, the Ramanujan-Nagell equation).
There is no hope for including 61 in the sequence but I found an interesting relation.
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Mickey1
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Re: Mersenne, Pell and Ramanujan end more
« Reply #1 on: Apr 18th, 2013, 12:38pm » |
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Let me add an interesting feature about Pell’s equation x(d)^2-dy(d)^2=1
If d is a prime then x(d-1) is also prime.
It doesn’t hold for all primes, but it is a pattern which holds for 17 of the 21 first (relevant) primes 2<d<100. (There are more primes below 100 but several values are irrelevant since x(d-1) does not exist d=2,5,17,37).
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