(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
I have NOT been able to find this problem in QUORA, so here goes my attempt towards a purported D1 solution on my own.
AT the outset, x+y divides both sides, so setting x+y=0, we have: (x,y)=(0,0) as the only nonnegative integer solution.
Now, dividing both sides of the given equation by x+y,, we have:
x+y=x(x-y) +y^2 .........(#)
Setting x=y=t(say), we have:
2t = t^2
=> t=0 or 2
=> (x,y) =(0,0), (2,2)
Now, comparing the coefficient of x in both sides of (#), we have:
x-y=1 and, y^2=y
This easily solves as: (x,y) = (2,1), (1,0)
Similarly, dividing the original equation by x+y,, we also have:
x+y = y(y-x)+x^2
So comparing the coefficient of y from both the sides, we have:
y-x=1 and x^2=x, which easily solves as:
(x,y) =(1,2), (0,1)
Consequently, (x,y) =(0,0), (0,1), (1,0), (1,2), (2,1) and (2,2) gives all possible nonnegative integer solutions to the problem under reference.
------------------------------------------------------------------------------------------------------------------------------
Thank you, AT.!
I thoroughly enjoyed SOLVING this puzzle and this problem is TRULY deserving of a D5 rating!!
Edited on December 29, 2021, 9:10 am
D1 Solution It Is!!!
