OFFSET
1,2
COMMENTS
Obviously all of x, y and z must be nonzero for all solutions N > 0. For any N = x*y*z = x+y+z, one gets -N from (-x, -y, -z), so considering only N >= 0 is not a restriction. Either none or exactly two among x, y and z must be negative.
For given N, the problem amounts to finding fractions x and y such that x*y^2 + x*(x - N)*y + N = 0, which in turn corresponds to finding rational points on the elliptic curve Y^2 = X^3 + N^2*(X+4)^2 (with X = -4*N/x and Y = 4*N*D/x^2, where D^2 is the discriminant of the previous quadratic in y).
It appears that (for N > 0) we have a rational solution iff this elliptic curve has nonzero rank. (Is there any counterexample?) If so, the sequence goes (0, 6, 7, 9, 13, 14, 15, 16, 19, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 37, 38, 40, 43, 44, 45, 46, 48, 49, 52, 53, 55, 56, 58, 59, 60, 61, 62, 63, 64, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 82, 83, 84, 86, 87, ...)
LINKS
Allan MacLeod, Elliptic Curves in Recreational Number Theory, arXiv:1610.03430 [math.NT], Oct. 2016.
Victor Miller and others, in reply to Keith F. Lynch, Re: Integer sums and products, math-fun mailing list (available for subscribers), Sep. 2024.
EXAMPLE
The first few terms correspond to the following solutions (|x| <= |y| <= |z|):
N | x | y | z
-----+---------+---------+---------
0 | 0 | 0 | 0 (or any rational y = -z).
6 | 1 | 2 | 3 (and also {25/21, 54/35, 49/15}).
7 | 7/6 | 4/3 | 9/2
9 | 1/2 | 4 | 9/2
13 | 36/77 | 121/42 | 637/66
14 | 1/3 | 9 | 14/3
15 | 1/2 | 5/2 | 12
16 | -2/3 | -4/3 | 18
19 | 121/234 | 324/143 |3211/198
...
All terms of A054000 (2*n^2-2: 0, 6, 16, 30, 48, 70, 96, 126, 160, 198, ...) are in the sequence, as product and sum of the triple (2*n^2, 1/n - 1, -1/n - 1).
PROG
(PARI) select( {is_A376243(n)=!n||ellrank(ellinit([0, 1, 0, 8, 16]*n^2))}, [0..30]) \\ Assuming there's a rational solution iff the elliptic curve has rank > 0. - M. F. Hasler, Sep 23 2024
CROSSREFS
KEYWORD
nonn,more
AUTHOR
M. F. Hasler, Sep 16 2024
STATUS
approved