A384101 a(n) is the least positive integer y such that y^2 = x^3 + x + n^2 for some positive integer x, or 0 if no such y exists.
0, 611, 265222, 23, 134316044, 1000300015, 292, 20662660277, 68722622488, 198364959099, 512009600030, 1207284678721, 2641831428132, 5429539323143, 44, 19683072900045, 18, 60717129072787, 182, 165216338968409, 262144307200060, 31, 618122334258242, 922190780558053
Offset: 0
Keywords
Examples
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) = 13*47 = 611. 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) = 2*132611 = 265222. (As for n = 1, this is the upper limit for a(n), given in FORMULA.) For n = 3, there is already x = 8 for which x^3 + x + 3^2 = 529 = 23^2 is a square, therefore a(3) = 23 is much smaller than the before mentioned upper limit.
Crossrefs
Cf. A384100 (the corresponding x-values).
Programs
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PARI
A384101(n, x=A384100(n))=sqrtint(n^2+x^3+x)
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Python
from math import isqrt from sympy.ntheory.primetest import is_square def A384101(n): if n==0: return 0 m = n**2 for x in range(1,n*(n**4*((n**4<<9)+96)+3)+1): if is_square(k:=x*(x**2+1)+m): return isqrt(k) # Chai Wah Wu, May 24 2025
Formula
0 < a(n) <= 2*n*(16*n^4 + 1)*(16*n^4 + 2) - n for all n > 0.
Extensions
a(9)-a(21) from Chai Wah Wu, May 24 2025
a(22)-a(23) from Jinyuan Wang, May 26 2025
Comments