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%I #28 Dec 24 2020 08:56:41
%S 1,2,3,6,91
%N All positive solutions, x, for each of two Diophantine equations noted by Richard K. Guy.
%C 2*(x^2)*((x^2)-1) = 3*((y^2)-1) has only these five positive solutions.
%C x*(x-1)/2 = (2^z)-1 has only these five positive solutions.
%C Richard K. Guy notes, as Example 29: "True, but why the coincidence?"
%C Algebraically, y solutions = {1, 3, 7, 29, 6761} can be derived from x solutions as follows: y = sqrt(((2*x^2 - 1)^2 + 5)/6). From this relationship it becomes clear that the form (((2*x^2 - 1)^2 + 5)/6) can only be an integer square for x is in {1, 2, 3, 6, 91}. Thus, x and y solutions are also unique integer solutions to the following equivalency: (2x^2 - 1)^2 = 6y^2 - 5. From this relationship the following statement naturally follows: ((sqrt(6*y^2 - 5) + 1)/2 - sqrt((sqrt(6*(y^2) - 5) + 1)/2))/2 = (2^z - 1) = {0, 1, 3, 15, 4095} = A076046(n), the Ramanujan-Nagell triangular numbers; z = {0, 1, 2, 4, 12} = (A060728(n) - 3). - _Raphie Frank_, Jun 26 2013
%H R. K. Guy, editor, <a href="/A339579/a339579.pdf">Western Number Theory Problems, 1985-12-21 & 23</a>, Typescript, Jul 13 1986, Dept. of Math. and Stat., Univ. Calgary, 11 pages. Annotated scan of pages 1, 3, 7, 9, with permission. See Problem 85:08.
%H Richard K. Guy, <a href="http://www.jstor.org/stable/2322249">The Strong Law of Small Numbers</a> (example #29).
%H R. K. Guy, <a href="/A005165/a005165.pdf">The strong law of small numbers</a>. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]
%F x = sqrt((sqrt(6*(y^2) - 5) + 1)/2) = (sqrt(2^(z + 3) - 7) + 1)/2; y = {1, 3, 7, 29, 6761} and z = (A060728(n) - 3) = A215795(n) = {0, 1, 2, 4, 12}. - _Raphie Frank_, Jun 23 2013
%Y Cf. A076046, A060728.
%K fini,full,nonn
%O 1,2
%A _Jonathan Vos Post_, Sep 05 2010
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