(x+y)^2=x^3+y^3
The above is a d2 or d3 problem if one tries simulation or algebraic solution.
Now solve it d1!
Non negative integers, of course !
Source: appeared as a question and an answer(super complicated!)in
QUORA
(In reply to
levels by Charlie)
(x+y)^2=x^3+y^3 can be factored into
(x+y) * (x^2-yx+y^2-x-y) = 0
Then by zero product property the graph is a union of the line y=-x and the conic curve x^2-yx+y^2-x-y = 0 The discriminant of the curve is (-1)^2-4*1*1=-3<0, so we know its an ellipse.
But I can manipulate the conic and rewrite it as (1/4)*(x+y-2)^2 + (3/4)(x-y)^2 = 1. x+y-2=0 and x-y=0 are orthogonal lines which make up the ellipse's two axis.
So taking x-y=0 and (3/4)(x-y)^2 = 1 gives (0,0) and (2,2) for the endpoints of the x-y=0 axis
Similarly taking x+y-2=0 and (1/4)*(x+y-2)^2 = 1 gives (1+sqrt(1/3), 1-sqrt(1/3)) and (1-sqrt(1/3), 1+sqrt(1/3)) for the endpoints of the x+y-2=0 axis.
Edited on December 30, 2021, 2:05 pm
re: levels - the graph
