OFFSET
0,2
COMMENTS
Otherwise said, first component of the lexicographically earliest positive integer solution (x, y) to x^3 + x + n^2 = y^2. See A384101 for the second component, y.
For any positive n, there is always the solution (x, y) = (8*n^2*(8*n^4 + 1), n*(512*n^8 + 96*n^4 + 3)). Therefore 0 < a(n) <= 8*n^2*(8*n^4 + 1) for all n > 0.
We remark that n = 3 and n = 6 are the only cases below n = 10 for which there is a smaller solution than S(n) = (x, y) given above, while gcd(x, y) = gcd(n, 3) (= 3 iff n is divisible by 3, otherwise 1).
LINKS
Rik Bos, How do I show that x^3 + x + a^2 = y^2 has at least one pair of positive integer solution (x, y) for each positive integer a? (Answer), Quora, May 12, 2025.
FORMULA
a(n) <= 8*n^2*(8*n^4 + 1) for all n > 0.
EXAMPLE
For n = 0, there can't be any positive x for which x^3 + x = x*(x^2 + 1) = y^2, therefore a(0) = 0. (Indeed, x^2 + 1 == 1 (mod x), so x has no factor in common with x^2 + 1 = y^2/x, so x must be a square itself, x = m^2. But then, x^2 + 1 = (y/m)^2 can't have a solution, since x^2 + 1 can't be a square.)
For n = 1, we can check that for x = 1, 2, 3, ..., value of x^3 + x + 1 = 3, 10, 31, ... isn't a square for any x < 72 which is the least positive integer so that x^3 + x + 1 = 72*(72^2 + 1) + 1 = 373321 = (13*47)^2 is a perfect square, thus a(1) = 72.
For n = 2, there is no x < 4128 for which x^3 + x + 2^2 is a square, but 4128*(4128^2+1) + 4 = (2*132611)^2 is indeed the least square of that form, so a(2) = 4128. (As for n = 1, this is the upper limit for a(n), given in FORMULA.)
For n = 3, there is a(3) = x = 8 for which x^3 + x + 3^2 = 529 = 23^2 is a square, much smaller than the upper limit for a(n).
PROG
(PARI) apply({A384100(n)=for(x=1, 64*n^6+8*n^2, issquare(n^2+x^3+x) && return(x))}, [0..6])
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, May 19 2025
EXTENSIONS
More terms from Jinyuan Wang, May 26 2025
STATUS
approved