mersenneforum.org Mersenne, Pell’s and the Ramanujan–Nagell equation
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 2013-03-03, 14:50 #1 Mickey1   Mar 2013 2 Posts Mersenne, Pell’s and the Ramanujan–Nagell equation 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 PE 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 Code: N d=2^n-3 d=2^(n-1)-3 PE 3 5 1 2^2-5*1^2=-1 4 13 5 8^2-13*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 (I had some difficulties with the spacing in the table). 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. We note that (2^n-4)^2/4 will obviously always be a natural number. 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). Last fiddled with by Batalov on 2013-03-03 at 20:18 Reason: formatted the table, added [code] tags

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