

A050788


Consider the Diophantine equation x^3 + y^3 = z^3  1 (x < y < z) or 'Fermat near misses'. Arrange solutions by increasing values of z. Sequence gives values of x.


5



6, 71, 135, 372, 426, 242, 566, 791, 236, 575, 1938, 2676, 1124, 2196, 1943, 1851, 1943, 7676, 3318, 10866, 3086, 3453, 17328, 4607, 28182, 10230, 25765, 31212, 7251, 34199, 6560, 15218, 29196, 54101, 32882, 51293, 17384, 8999, 58462, 75263
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OFFSET

1,1


COMMENTS

From Fred W. Helenius (fredh(AT)ix.netcom.com), Jul 22 2008: (Start)
There is an infinite family of solutions to c^3+1=a^3+b^3 given by
(a,b,c) = (9n^3 + 1, 9n^4, 9n^4 + 3n). The present sequence actually asks about
x^3 + y^3 = z^3  1 with x < y < z; for that we can take
(x,y,z) = (9n^3  1, 9n^4  3n, 9n^4) for n > 1.
I extracted these solutions from Theorem 235 in Hardy & Wright; the result shown there is that all nontrivial rational solutions of
x^3 + y^3 = u^3 + v^3 are given by
x = r(1  (a  3b)(a^2 + 3b^2))
y = r((a + 3b)(a^2 + 3b^2)  1)
u = r((a + 3b)  (a^2 + 3b^2)^2)
v = r((a^2 + 3b^2)^2  (a  3b))
where r,a,b are rational and r is not zero.
Specializing to r = 1, b = n/2 and a = 3n/2 gives
x = 1, y = 9n^3  1, u = 3n  9n^4, v = 9n^4.
The solutions given above are obtained by changing signs and moving cubes from one side of the equation to the other as necessary.
Unfortunately, not all integral solutions are found so easily: the third value in A050788 corresponds to 135^3 + 138^3 = 172^3  1; this is not produced by such simple choices of r,a,b. (End)


REFERENCES

Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107124.
David Wells, "Curious and Interesting Numbers", Revised Ed. 1997, Penguin Books, On number "729", p. 147.


LINKS



EXAMPLE

(575)^3 + 2292^3 = 2304^3  1.


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



