

A322548


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


0




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



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).


KEYWORD

nonn,fini,full


AUTHOR



STATUS

approved



