BdMO Online Forum

Anindya Biswas wrote: ↑

Mon Jan 11, 2021 1:59 pm

Find all integers $x,y$ such that $x^3+y^3=(x+y)^2$

$\textbf{Claim:}$ The equation is only true for non-negative integers.
Proof : Assume that both of the integers are negative. Hence the right side is positive, where the left side is negative, which is not possible.
Now, Assume that one of them is negative. WLOG assume that, y is negative , we have, $x^3-y^3=(x-y)^2 \Rightarrow x^2+xy+y^2=x-y $,
$x^2+xy+y^2 > x^2 > x > x-y \Rightarrow x^2+xy+y^2 \neq x-y$

Now, assume that one of them is $0$. Then the only solution is $(x,y)=(0,0),(1,0),(0,1)$

Now, assume that $x$, $y$ are positive integers, Let $gcd(x,y)=g$, Assume $x=ga,y=gb$(Where $a$ and $b$ are coprime)
so we have,
$g(a^3+b^3)=(a+b)^2 $
$\Rightarrow g(a^2-ab+b^2)=(a+b) $
$\Rightarrow g = \dfrac{a+b}{a^2-ab+b^2}$
$\Rightarrow a+b \geq a^2-ab+b^2$
$\Rightarrow a+b+ab\geq a^2+b^2 \geq 2ab$
$\Rightarrow a+b \geq ab$
$\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} \geq 1$
It's easy to see that the only integers satifying the inequality is $(a,b)=(2,2),(n,1),(1,n)$(where $n \in \mathbb{N}$)
Again we have, $n+1 \geq n^2-n+1 \Rightarrow 1 \geq (n-1)^2 \Rightarrow n=1,2$
So plugging these into the original equation , we have, $(x,y)=(2,1),(1,2),(2,2)$
So, all the solutions to the equation is $(x,y)=(0,0),(1,0),(2,1),(2,2)$ and all its permutations.$\blacksquare$
[sorry if there's any mistake. ]

~Aurn0b~ wrote: ↑

Thu Jan 14, 2021 11:35 pm

Anindya Biswas wrote: ↑

Mon Jan 11, 2021 1:59 pm

Find all integers $x,y$ such that $x^3+y^3=(x+y)^2$

$\textbf{Solution :}$

$\textbf{Claim:}$ The equation is only true for non-negative integers.
Proof : Assume that both of the integers are negative. Hence the right side is positive, where the left side is negative, which is not possible.
Now, Assume that one of them is negative. WLOG assume that, y is negative , we have, $x^3-y^3=(x-y)^2 \Rightarrow x^2+xy+y^2=x-y $,
$x^2+xy+y^2 > x^2 > x > x-y \Rightarrow x^2+xy+y^2 \neq x-y$

Now, assume that one of them is $0$. Then the only solution is $(x,y)=(0,0),(1,0),(0,1)$

Now, assume that $x$, $y$ are positive integers, Let $gcd(x,y)=g$, Assume $x=ga,y=gb$(Where $a$ and $b$ are coprime)
so we have,
$g(a^3+b^3)=(a+b)^2 $
$\Rightarrow g(a^2-ab+b^2)=(a+b) $
$\Rightarrow g = \dfrac{a+b}{a^2-ab+b^2}$
$\Rightarrow a+b \geq a^2-ab+b^2$
$\Rightarrow a+b+ab\geq a^2+b^2 \geq 2ab$
$\Rightarrow a+b \geq ab$
$\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} \geq 1$
It's easy to see that the only integers satifying the inequality is $(a,b)=(2,2),(n,1),(1,n)$(where $n \in \mathbb{N}$)
Again we have, $n+1 \geq n^2-n+1 \Rightarrow 1 \geq (n-1)^2 \Rightarrow n=1,2$
So plugging these into the original equation , we have, $(x,y)=(2,1),(1,2),(2,2)$
So, all the solutions to the equation is $(x,y)=(0,0),(1,0),(2,1),(2,2)$ and all its permutations.$\blacksquare$
[sorry if there's any mistake. ]

Anindya Biswas wrote: ↑

Sun Jan 17, 2021 7:11 pm

~Aurn0b~ wrote: ↑

Thu Jan 14, 2021 11:35 pm

Anindya Biswas wrote: ↑

Mon Jan 11, 2021 1:59 pm

Find all integers $x,y$ such that $x^3+y^3=(x+y)^2$

$\textbf{Solution :}$

$\textbf{Claim:}$ The equation is only true for non-negative integers.
Proof : Assume that both of the integers are negative. Hence the right side is positive, where the left side is negative, which is not possible.
Now, Assume that one of them is negative. WLOG assume that, y is negative , we have, $x^3-y^3=(x-y)^2 \Rightarrow x^2+xy+y^2=x-y $,
$x^2+xy+y^2 > x^2 > x > x-y \Rightarrow x^2+xy+y^2 \neq x-y$

Now, assume that one of them is $0$. Then the only solution is $(x,y)=(0,0),(1,0),(0,1)$

Now, assume that $x$, $y$ are positive integers, Let $gcd(x,y)=g$, Assume $x=ga,y=gb$(Where $a$ and $b$ are coprime)
so we have,
$g(a^3+b^3)=(a+b)^2 $
$\Rightarrow g(a^2-ab+b^2)=(a+b) $
$\Rightarrow g = \dfrac{a+b}{a^2-ab+b^2}$
$\Rightarrow a+b \geq a^2-ab+b^2$
$\Rightarrow a+b+ab\geq a^2+b^2 \geq 2ab$
$\Rightarrow a+b \geq ab$
$\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} \geq 1$
It's easy to see that the only integers satifying the inequality is $(a,b)=(2,2),(n,1),(1,n)$(where $n \in \mathbb{N}$)
Again we have, $n+1 \geq n^2-n+1 \Rightarrow 1 \geq (n-1)^2 \Rightarrow n=1,2$
So plugging these into the original equation , we have, $(x,y)=(2,1),(1,2),(2,2)$
So, all the solutions to the equation is $(x,y)=(0,0),(1,0),(2,1),(2,2)$ and all its permutations.$\blacksquare$
[sorry if there's any mistake. ]

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...
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 $y$.