A180445 All positive solutions, x, for each of two Diophantine equations noted by Richard K. Guy.

1, 2, 3, 6, 91

Offset

1,2

Comments

2*(x^2)*((x^2)-1) = 3*((y^2)-1) has only these five positive solutions.

x*(x-1)/2 = (2^z)-1 has only these five positive solutions.

Richard K. Guy notes, as Example 29: "True, but why the coincidence?"

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

Links

Table of n, a(n) for n=1..5.

R. K. Guy, editor, Western Number Theory Problems, 1985-12-21 & 23, 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.

Richard K. Guy, The Strong Law of Small Numbers (example #29).

R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]

Formula

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

Crossrefs

Cf. A076046, A060728.

Sequence in context: A018474 A018488 A018506 * A018531 A018546 A018553

Adjacent sequences: A180442 A180443 A180444 * A180446 A180447 A180448

Keyword

fini,full,nonn

Author

Jonathan Vos Post, Sep 05 2010

Status

approved