Cute little Diophantine!!
Posted: Mon Jan 11, 2021 1:59 pm
Find all integers $x,y$ such that $x^3+y^3=(x+y)^2$
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$\textbf{Solution :}$
One thing I love about diophantine equation is the variety of ways one can solve them. Your consideration of gcd is pretty impressive and the way you manipulated those inequalities. But there's a little mistake you have made and some solutions are missing...
Ah!...Sorry for the mistake, Thank you for pointing that out.Anindya Biswas wrote: ↑Sun Jan 17, 2021 7:11 pmOne thing I love about diophantine equation is the variety of ways one can solve them. Your consideration of gcd is pretty impressive and the way you manipulated those inequalities. But there's a little mistake you have made and some solutions are missing...
The first claim isn't true, there are negative integers satisfying this equation. The part where you wrote $x^3-y^3=(x-y)^2\Rightarrow x^2+xy+y^2=x-y$, you have missed a case where $x-y=0$. That case gives infinitely many pairs of integers $(x,y)=(n, -n)$ as solutions. The part where you got all the solutions where $x,y\geq0$ is totally fine.
When I solved it, I didn't considered their gcd, instead I considered the equation as a quadratic equation with variable $x$ and used the inequality for discriminant to get a bound on $y$.