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A103268
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Positive integers x such that there exist positive integers y and z satisfying x^3 + y^3 = z^5.
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1
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OFFSET
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1,1
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COMMENTS
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Warning! These terms have not been proved to be correct. There may be missing terms. - N. J. A. Sloane, Jan 05 2007
There are no solutions with (x,y,z) relatively prime. [Bruin]
Trivially, if m^3 + n^3 = z^2, then (z*m)^3 + (z*n)^3 = z^5. So from A103254 we can find many solutions. - James Mc Laughlin, Jan 30 2007
For max(x,y) < 1.1*10^12, there are no more terms < 1458. Most likely this is true for all x,y. - Chai Wah Wu, Jan 15 2016
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LINKS
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Nils Bruin, On powers as sums of two cubes, in Algorithmic number theory (Leiden, 2000), 169-184, Lecture Notes in Comput. Sci., 1838, Springer, Berlin, 2000.
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EXAMPLE
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1 + 2^3 = 3^2 so 3^3 + 6^3 = 3^5 and 3 and 6 are terms.
With max(x,y) < 10^4, we have these [x,y,z] triples: [3, 6, 3], [8, 8, 4], [96, 192, 24], [256, 256, 32], [729, 1458, 81], [1944, 1944, 108], [686, 2058, 98], [3696, 4368, 168], [3072, 6144, 192], [8192, 8192, 256], [2508, 8436, 228], ... - David Broadhurst, Jan 30 2007
These are variously immediate consequences of 1 + 1 = 2, 1 + 2^3 = 3^2, 1 + 3^3 = 2^2*7 and, much more unexpectedly, 11^3 + 37^3 = 2^4*3^2*19^2. The last example shows that solutions with a common factor are not completely trivial. [Comment based on email from Alf van der Poorten, Feb 15 2007]
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MATHEMATICA
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r[z_] := Reduce[x > 0 && y > 0 && x^3 + y^3 == z^5, {x, y}, Integers];
sols = Reap[Do[rz = r[z]; If[rz =!= False, xyz = {x, y, z} /. {ToRules[rz]}; Print[xyz]; Sow[xyz]], {z, 1, 1000}]][[2, 1]] // Flatten[#, 1]&;
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CROSSREFS
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KEYWORD
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more,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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