From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math.num-analysis,sci.math Subject: Re: Sum of two powers = Square Date: 8 Jan 1998 22:13:19 GMT In article <34B520CD.B748A1F2@hotmail.com>, Dario Alejandro Alpern writes: |> An open question: The set of a^4 + b^3 = c^2 with gcd(a,b,c)=1 is finite |> or infinite? Infinite. The Pell equation 3 x^2 + 1 = a^2 has an infinite number of positive integer solutions. If b = x^2 - 1 and c = a^2 + (x-1)^3, then a^4 + b^3 - c^2 = 2 (x-1)^3 (3 x^2 + 1 - a^2) = 0 When x is odd a,b,c will all be divisible by 2, but otherwise gcd(a,b)=1. The first few nontrivial cases in this family are [7, 15, 76], [26, 224, 3420], [97, 3135, 175784], [362, 43680, 9129956], [1351, 608399, 474554340], [5042, 8473920, 24667592764], [18817, 118026495, 1282239885136], [70226, 1643897024, 66651805000260] Another family comes from the integer solutions of 3 x^2 = 2 y^2 + 1, with a = y (1 + 3 x + 3 x^2), b = x (1 + 2 x) (2 + 3 x), c = a^2 + (1+2 x)^3. The first few of these are [7, 15, 76], [2981, 4959, 8893220], [2619379, 4285439, 6861152080980] Robert Israel israel@math.ubc.ca Department of Mathematics (604) 822-3629 University of British Columbia fax 822-6074 Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: ksbrown@seanet.com (Kevin Brown) Newsgroups: sci.math.num-analysis,sci.math Subject: Re: Sum of two powers = Square Date: Fri, 09 Jan 1998 01:06:18 GMT Dario Alejandro Alpern wrote: > These are the sums that I found using the UBASIC program listed in my > post of Nov 20th, 1997 (search for it in http://www.dejanews.com): > 1 ^ n + 2 ^ 3 = 3 ^ 2 > 5 ^ 4 + 6 ^ 3 = 29 ^ 2 > 7 ^ 4 + 15 ^ 3 = 76 ^ 2 > 17 ^ 4 + 42 ^ 3 = 397 ^ 2 > 97 ^ 4 + 3135 ^ 3 = 175784 ^ 2 > [snip] > Notice that the gcd of the three bases are 1. An open question: The > set of a^4 + b^3 = c^2 with gcd(a,b,c)=1 is finite or infinite? Infinite. To see this, factor the equation b^3 = c^2 - a^4 as b^3 = (c + a^2)(c - a^2) If a and c have opposite parity the two factors on the right are coprime, so we have integers m,n such that m^3 = c + a^2 n^3 = c - a^2 It follows that we have a solution if we can find integers a,c of the form c = (m^3 + n^3)/2 a^2 = (m^3 - n^3)/2 These values of a^2 and c are clearly integers (since m and n are both odd), so the only constraint on m,n is that they make "a" an integer. There are several ways of constructing an infinite family with this property. For example, if we restrict ourselves to "twin" values of m and n, meaning that m=k+1 and n=k-1, the above equation for a^2 becomes a^2 - 3k^2 = 1 which is a "Pell equation". By the usual methods we have the solutions given by k = 0, 1, 4, 15, 56, 209, 780, ... and in general by the recurrence k_j = 4 k_{j-1} - k_{j-2} Obviously these values of k alternate between odd and even. The odd values give even values of m and n, so those correspond to non- primitive solutions, such as 26^4 + 224^3 = 3420^2, whereas the even values of k give odd values of m and n, and the corresponding solutions of the original equation are primitive. For example, with k=4 we have m=5, n=3, which gives the solution a=7,b=15,c=76. The next primitive solution in this family is based on the Pell solution k=56, which gives m=57,n=55 and so a=97,b=3135,c=175784. The next such solution would be based on the even Pell solution k=780, and so on. We can create other infinite families in a similar way by defining a linear relation between m and n. ________________________________________________________ | MathPages /*\ http://www.seanet.com/~ksbrown/ | | / \ | |__________/"Nothing remains but the dull cohesion _____| of old surfaces too weak to disolve."