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A173195
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Values of n such that 4^x + 4^y + 4^z = n^2 with arbitrary integers x <= y <= z.
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2
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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
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OFFSET
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1,1
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COMMENTS
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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.
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REFERENCES
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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.
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LINKS
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FORMULA
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n = 2^x + 2^(2y-x-1), and z = 2y - x - 1.
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EXAMPLE
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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
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MAPLE
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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 :
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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