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 A322548 Integers x such that x^2 + 119 = 15*2^y. 0
 1, 11, 19, 29, 61, 701 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The exponents y of the corresponding powers of 2 are 3, 4, 5, 6, 8, 15. The list gives all positive integers x such that x^2 + 119 = 15*2^y. Yann Bugeaud proposed the problem to prove that there is an absolute constant C such that, for any positive integers D, k and a prime number p such that gcd(D, kp) = 1, the Diophantine equation x^2 + D = k*p^n has at most C integer solutions (x, n) (Problem 9 of the list of 22 open problems below). LINKS Table of n, a(n) for n=1..6. Jan-Hendrik Evertse, Some open problems about Diophantine equations, 22 problems posed at the Instructional conference and workshop "Solvability of Diophantine equations", May 7-16, 2007, Lorentz Center, Leiden. Jörg Stiller, The Diophantine equation x^2 + 119 = 15*2^n has exactly six solutions, Rocky Mountain J. Math. 26 (1996), 295-298. EXAMPLE a(2) = 11: 11^2 + 119 = 240 = 15*2^4. MATHEMATICA s={}; Do[r = Solve[x^2 + 119 == 15*2^k && x >= 0, x, Integers]; If[Length[r]>0, AppendTo[s, x/.r[[1]]]], {k, 1, 15}]; s (* Amiram Eldar, Dec 15 2018 *) CROSSREFS Cf. A038198 (All solutions to x^2 + 7 = 2^y). Sequence in context: A175275 A094517 A196669 * A049719 A155555 A357426 Adjacent sequences: A322545 A322546 A322547 * A322549 A322550 A322551 KEYWORD nonn,fini,full AUTHOR Tomohiro Yamada, Dec 14 2018 STATUS approved

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