(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
Factor each side: (x+y)*(x+y) = (x+y)*(x^2-xy+y^2)
Now either x+y=0 or x+y=x^2-xy+y^2,
If x+y=0 then in nonnegative integers the only solution is (0,0).
Then if x+y=x^2-xy+y^2, subtract xy from each side and simplify:
x+y-xy = (x-y)^2.
This implies x+y-xy >= 0. Then we want a pair of nonnegative integers whose sum is greater or equal to their product.
If x and y are both at least 2 then to satisfy x+y-xy >= 0 they must be (2,2). Checking in x+y=x^2-xy+y^2 shows that (2,2) is a solution.
Then the remaining cases are when one of x,y is 0 or 1. Wlog, let x be 0 or 1.
If x=0 then x+y=x^2-xy+y^2 reduces to y = y^2, which makes y=0 or 1. Then (0,1) and its reflection (1,0) must be solutions.
If x=1 then x+y=x^2-xy+y^2 reduces to 1+y=1-y+y^2, which has y=0 or 2 as roots. Then (1,2) and its reflection (2,1) must be solutions.
So in total there are six solutions in nonnegative integers: (0,0), (0,1), (1,0), (1,2), (2,1), (2,2).
This is a LOT of work for a problem that Ady refers to as Level 1 difficulty. I'd put it closer to difficulty 2.
Solution
