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A135509
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Nonnegative integers c such that there are nonnegative integers a and b that satisfy a^(1/2) + b^(1/2) = c^(1/2) and a^2 + b = c.
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1
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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
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OFFSET
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0,3
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COMMENTS
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Define FLTR as Fermat's Last Theorem with rational exponents. Consider x + y = x + y. Then (x^m)^(1/m) + (y^m)^(1/m) = ((x+y)^m)^(1/m) for integer m >= 1.
For m = 2, we have (x^2)^(1/2) + (y^2)^(1/2) = ((x+y)^2)^(1/2). Thus, 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 + 2*x*y + y^2 => y = (x^3-x)/2. It follows that c = (x+y)^2 = (x^3 + x)^2/4 is the generating function for this sequence for x = 0, 1, 2, 3, ...
For m = 2, there are infinitely many nonnegative integer solutions for the FLTR proposition. The same holds for m = 3, i.e., there are also infinitely many nonnegative integer solutions to a^(1/3) + b^(1/3) = c^(1/3). E.g., 8^(1/3) + 27^(1/3) = 125^(1/3). Moreover, there are infinitely many solutions to FLTR for a general positive integer m.
However, in conjunction with a^2 + b = c, I could not find any nontrivial solutions when m >= 3. Perhaps there is another formula that will yield solutions. [Edited by Petros Hadjicostas, Dec 21 2019]
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LINKS
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FORMULA
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a(n) = (n + n^3)^2/4 = A006003(n)^2.
G.f.: x*(1 + x)*(1 + 4*x + x^2)*(1 + 13*x + x^2)/(1 - x)^7. (End)
E.g.f.: (1/4)*x*(4 + 46*x + 102*x^2 + 67*x^3 + 15*x^4 + x^5)*exp(x). - G. C. Greubel, Oct 16 2016
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EXAMPLE
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For a = 9, b = 144, and c = 225, we obtain 9^(1/2) + 144^(1/2) = 225^(1/2) and 9^2 + 144 = 225. Thus, c = 225 is an entry in this sequence.
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MATHEMATICA
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Table[(n + n^3)^2/4, {n, 0, 25}] (* G. C. Greubel, Oct 16 2016 *)
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PROG
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(PARI) flt2(n) = {local(a, b); for(a=0, n, b = (a^3+a)/2; print1(b^2", "))} /* edited by Petros Hadjicostas, Dec 21 2019 */
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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