A173195 Values of k such that 4^x + 4^y + 4^z = k^2 with arbitrary integers x <= y <= z.

3, 6, 9, 12, 18, 24, 33, 36, 48, 66, 72, 96, 129, 132, 144, 192, 258, 264, 288, 384, 513, 516, 528, 576, 768, 1026, 1032, 1056, 1152, 1536, 2049, 2052, 2064, 2112, 2304, 3072, 4098, 4104, 4128, 4224, 4608, 6144, 8193, 8196, 8208, 8256, 8448, 9216, 12288

Offset: 1

Author

Michel Lagneau, Feb 12 2010

Keywords

• nonn
• easy

Comments

We prove that the solutions of 4^x + 4^y + 4^z = k^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 = k^2.

From Frederik P.J. Vandecasteele, Jun 06 2025: (Start)

For a given n, the exponents are x = A384688(n-1), y = A138099(n), z = A000267(n-1) so that a(n) = 2^A384688(n-1) + 2^A000267(n-1).

Terms are all and only those k whose binary expansion is two 1 bits an odd distance apart. (End)

Examples

			x = 0, y = 1 then z = 1, and k = 3.
x = 1, y = 2 then z = 2, and k = 6.
x = 0, y = 2 then z = 3, and k = 9.
		

References

• T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.
• J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Ellipses, 2004.
• H. N. Shapiro, Introduction to the Theory of Numbers, John Wiley & Sons, 1983.

Links

• David A. Corneth, Table of n, a(n) for n = 1..9999 (terms <= 10^60)

Crossrefs

Cf. A263132.

Subsequence of A018900.

Cf. A384688, A138099, A000267.

Programs

Maple

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 :

Mathematica

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 *)

Formula

k = 2^x + 2^(2y-x-1), and z = 2y - x - 1.

Conjecture: a(n) = 3*A263132(n). - George Beck, May 05 2021