%I #21 May 05 2021 18:15:36

%S 3,6,9,12,18,24,33,36,48,66,72,96,129,132,144,192,258,264,288,384,513,

%T 516,528,576,768,1026,1032,1056,1152,1536,2049,2052,2064,2112,2304,

%U 3072,4098,4104,4128,4224,4608,6144,8193,8196,8208,8256,8448,9216,12288

%N Values of n such that 4^x + 4^y + 4^z = n^2 with arbitrary integers x <= y <= z.

%C We prove that the solutions of 4^x + 4^y + 4^z = n^2 are (x,y,2y-x-1), for any arbitrary integer x,y. We calculate z. 4^x + 4^y + 4^z is square if positive integers m and odd integer t are such as : 1 + 4^(y-x) + 4^(z-x) = (1 + t.2^m)^2, that's why : (1 + 4^(z-y) .( 4^(y-x)) = t(1 + t.2^(m+1)) t.2^(m+1), and then m = 2y - 2x - 1. If we report this value in the precedent equation, we obtain : t-1 = (2^(z-2y+x+1) + t)(2^(z-2y+x+1) - t) . 4^(y-x-1). Because t is odd, z = 2y - x - 1. Finally, this values gives the square (2^x + 2^(2y-x-1))^2 = n^2.

%D T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.

%D J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Ellipses, 2004.

%D H. N. Shapiro, Introduction to the Theory of Numbers, John Wiley & Sons, 1983.

%H David A. Corneth, <a href="/A173195/b173195.txt">Table of n, a(n) for n = 1..9999</a> (terms <= 10^60)

%F n = 2^x + 2^(2y-x-1), and z = 2y - x - 1.

%F Conjecture: a(n) = 3*A263132(n). - _George Beck_, May 05 2021

%e x=0, y=0 then z = 1, and n = 3 x=1, y = 2 then z=2, and n = 6 x=0, y=y then z = 3, and n = 9

%p for x from 0 to 1000 do :for y from x to 1000 do: n := evalf(2^x + 2^(2*y-x-1)): print (n) ; od :od :

%t Take[Union[Select[Sqrt[Flatten[Table[(2^x + 2^(2*y - x - 1))^2, {x, 0, 13}, {y, 0, 13}]]], IntegerQ]],49] (* _Jean-François Alcover_, Sep 13 2011 *)

%Y Cf. A263132.

%K nonn

%O 1,1

%A _Michel Lagneau_, Feb 12 2010