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KISS solution (Posted on 2021-12-29) Difficulty: 1 of 5
(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

See The Solution Submitted by Ady TZIDON    
Rating: 5.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution levels | Comment 2 of 8 |
Depending on one's degree of insight:

Level 1 would see that x = y = 0 would satisfy the equality.

Level 2 would find any y = -x would satisfy, with both sides being zero.

Level 3 could find 2,2 with both sides being 16.

Level 4 could find 1,2 and 2,1 with both sides = 9.

It's really made simple by leaving the grunt work to the computer to find:

1. The straight line y = -x

2. Something that looks visually like an ellipse, with a major axis from 0,0 to 2,2 and minor axis from 1,2 to 2,1.

That's wrong; the minor axis is from (1 + 1/sqrt(3),  1 - 1/sqrt(3)) to (1 - 1/sqrt(3),  1 + 1/sqrt(3)). What was I thinking?


clearvars, clc
for x=-1:.01:2.1
    y1 = (1/2)*(sqrt(-3*x^2 + 6*x + 1) + x + 1);
    y2 = (1/2)*(-sqrt(-3*x^2 + 6*x + 1) + x + 1);
    disp([x y1 y2])
end
syms x y
eq=(x+y)^2==x^3+y^3
fimplicit(eq)

after finding those y values, in terms of x, with Wolfram Alpha.

Edited on December 29, 2021, 3:16 pm
  Posted by Charlie on 2021-12-29 14:55:05

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