OFFSET
1,1
COMMENTS
With z1 = 0, z2 = x+i*y, z3 = x-i*y and P(z) = (z-z1)*(z-z2)*(z-z3) we obtain:
(1) P(z) = z^3 - 2*x*z^2 + (x^2+y^2)*z.
(2) P'(z) = 3*z^2 - 4*x*z + x^2+y^2.
The two zeros of P'(z) are: (2*x-sqrt(x^2-3*y^2))/3 and (2*x+sqrt(x^2-3*y^2))/3.
REFERENCES
Beniamin Bogosel, A Geometric Proof of the Siebeck-Marden Theorem, Amer. Math. Monthly, vol. 125, no 4, 2017, p. 459-463.
A. Eydelzon, On a New Property of the Steiner Inellipse, Amer. Math. Monthly, vol. 127, no 10, 2020, p. 933-935.
LINKS
EXAMPLE
(a(1),a(2)) = (3,5) because from (2) with (x,y) = (6,3), the two zeros of P'(z) are: (2*6-sqrt(6^2-3*3^2))/3 = (12 - sqrt(9))/3 = 3 and (2*6+sqrt(6^2-3*3^2))/3 = (12 + sqrt(9))/3 = 5. The two foci are integers.
(a(7),a(8)) = (13,15) because from (2) with (x,y) = (21,12), the two zeros of P'(z) are (2*21-sqrt(21^2-3*12^2))/3 = (42 - sqrt(9))/3 = 13 and (2*21+sqrt(21^2-3*12^2))/3 = (12 + sqrt(9))/3 = 15. The two foci are integers.
MAPLE
nn:=200:
for x from 1 to nn do:
for y from 1 to nn while(x^2>3*y^2) do:
u:=sqrt(x^2+y^2):v:=2*b:s:=sqrt(x^2+y^2)+y:
z1:=(2*x-sqrt(x^2-3*y^2))/3:z2:=(2*x+sqrt(x^2-3*y^2))/3:
if z1=floor(z1) and z2=floor(z2) then printf(`%d, `, z1):
printf(`%d, `, z2):
else fi:
od:
od:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 16 2025
STATUS
approved
