A322548 Integers x such that x^2 + 119 = 15*2^y.

1, 11, 19, 29, 61, 701

Offset: 1

Author

Tomohiro Yamada, Dec 14 2018

Keywords

• nonn
• fini
• full

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).

Examples

			a(2) = 11: 11^2 + 119 = 240 = 15*2^4.
		

Links

• 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.

Crossrefs

Cf. A038198 (All solutions to x^2 + 7 = 2^y).

Programs

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 *)