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A135509
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Values c^(1/2) from a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c.
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0
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0, 1, 25, 225, 1156, 4225, 12321, 30625, 67600, 136161, 255025, 450241, 756900, 1221025, 1901641, 2873025, 4227136, 6076225, 8555625, 11826721, 16080100, 21538881, 28462225, 37149025, 47941776, 61230625, 77457601, 97121025
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| For n=2, there is an infinite number of solutions for the proposition. For n > 2, there is an infinite number of solutions to a^(1/3) + b^(1/3) = c^(1/3). Moreover, there is an infinite number of solutions to FLTR for general n. E.g. 8^(1/3) + 27^(1/3) = 125^(1/3). However, in conjunction with a^2+b = c, I could not find any nontrivial solutions. Perhaps there is another formula that will yield solutions.
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FORMULA
| Define FLTR as Fermat's Last Theorem with rational exponents. Consider x + y = x + y. Then (x^n)^(1/n) + (y^n)^(1/n) = ((x+y)^n)^(1/n) or FLTR For n=2 we have, (x^2)^(1/2) + (y^2)^(1/2) = ((x+y)^2)^(1/2) So a = x^2, b = y^2 and c = (x+y)^2 Then a^2 + b = c => x^4 + y^2 = (x+y)^2 x^4 + y^2 = x^2 + 2xy + y^2 y = (x^3-x)/2 (x+y)^2 is the generating function for this sequence for x=0,1,2,3,..
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EXAMPLE
| For a=9,b=144,c=225, 9^(1/2) + 144^(1/2) = 225^(1/2) and 9^2 + 144 = 225. So
c = 225 is the 4th entry in the sequence.
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PROG
| (PARI) flt2(n, p) = { local(a, b); for(a=0, n, b = (a^3-a)/2; print1(b^2", ") ) }
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CROSSREFS
| Sequence in context: A030484 A038692 A017330 * A065779 A095241 A061839
Adjacent sequences: A135506 A135507 A135508 * A135510 A135511 A135512
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KEYWORD
| nonn
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AUTHOR
| Cino Hilliard (hillcino368(AT)hotmail.com), Feb 09 2008
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