A348139 Three-digit numbers abc such that the quadratic equation ax^2 + bx + c = 0 has a rational root.

100, 110, 120, 121, 130, 132, 140, 143, 144, 150, 154, 156, 160, 165, 168, 169, 170, 176, 180, 187, 190, 198, 200, 210, 220, 230, 231, 240, 242, 250, 252, 253, 260, 264, 270, 273, 275, 276, 280, 286, 288, 290, 294, 297, 299, 300, 310, 320, 330, 340, 341, 350, 352, 360, 363, 370, 372, 374, 380, 384, 385, 390, 396, 400, 410, 420, 430, 440, 441, 450, 451, 460, 462, 470

Offset

1,1

Comments

Inequalities: 1 <= a <= 9, 0 <= b, c <= 9.

If the quadratic equation ax^2 + bx + c = 0 has a rational root, then b^2-4ac is a square, the two roots are rational and nonpositive.

Proposition: these three-digit numbers abc are all composite.

The Olympiad problem proposed in Changhua, Taiwan, 2010 (see Reference) asked for a proof that the three-digit number abc is not a prime number.

If abc is a term with a, b, c >= 1 then cba is another term.

The total number of terms is 147.

The first 19 terms are also the first 19 terms of A033828, then A033828(20) = 182 while a(20) = 187.

Also, the first 23 terms are the first 23 3-digit terms of A267509, from A267509(39) to A267509(61), then A267509(62) = 202 while a(24) = 210.

References

Xiong Bin and Lee Peng Yee, Mathematical Olympiad in China (2009-2010), Problems and Solutions, Changhua, Taiwan, 2010, First Day, Problem 1, p. 147, East China Normal university Press - World Scientific, 2013.

Links

Michel Marcus, Table of n, a(n) for n = 1..147

Index to sequences related to Olympiads.

Example

x^2 + 2x = x*(x+2), whose roots are {-2, 0}, so 120 is a term.
2x^2 = 0 has double root {0}, so 200 is a term.
4x^2 + 7x + 3 = 4*(x+1)*(x+3/4), whose roots are {-3/4, -1}, so 473 = 11*43 is a term.

Mathematica

Select[Range[100, 999], (d = (#[[2]]^2 - 4*#[[1]]*#[[3]])&@ IntegerDigits[#]) >= 0 && IntegerQ @ Sqrt[d] &] (* Amiram Eldar, Oct 02 2021 *)

Prog

(Python)
from math import isqrt
def ok(n):
    s = str(n)
    if len(s) != 3: return False
    a, b, c = list(map(int, s))
    D = b**2 - 4*a*c
    return D >= 0 and isqrt(D)**2 == D
def afull(): return [m for m in range(100, 1000) if ok(m)]
print(afull()) # Michael S. Branicky, Oct 02 2021
(PARI) isok(m) = my(d=digits(m)); (#d==3) && issquare(d[2]^2 - 4*d[1]*d[3]); \\ Michel Marcus, Oct 03 2021

Crossrefs

Cf. A033828, A267509.

Sequence in context: A044866 A290948 A033828 * A162530 A204590 A115454

Adjacent sequences: A348136 A348137 A348138 * A348140 A348141 A348142

Keyword

nonn,base,fini,full

Author

Bernard Schott, Oct 02 2021

Status

approved