Solve in $\mathbb Z$ the equation $(x+y)(x^2+y^2)=4xy+3$

Question

So I have $$(x+y)(x^2+y^2)=4xy+3$$ I tried to develop it and I got $$x^3+y^3+xy(x+y-4)=3$$ what can I do next?

Answer 1

Not an answer, but a thing that is too big for a comment.

I run some Mathematica code to look for solutions to your problem, using the following code:

In[1]:=FullSimplify[
 Solve[{(x + y)*(x^2 + y^2) == 
    4*x*y + 3, -10^5 <= x <= 10^5 && -10^5 <= y <= 10^5}, {x, y}, 
  Integers]]

That code looks for answers on your question between the boundaries $-10^5\le x\le10^5$ and $-10^5\le\text{y}\le10^5$ and it gave me:

Out[1]={{x -> -2, y -> 1}, {x -> 1, y -> -2}}

So, there are $2$ solutions in the bounds $-10^5\le x\le10^5$ and $-10^5\le\text{y}\le10^5$.

Answer 2

$$(x+y)(x^2+y^2)=4xy+3\tag{1}$$ Let $X=x+y, Y=xy$ then we get $$X^3-2XY-4Y-3 = 0\tag{2}$$
Hence $Y=1/2(X^3-3)/(X+2)$
We can get ${x,y}$ by solving for $t^2-Xt+Y=0.$
Since discriminant must be positive, we know that $-2\leqq X\leqq2.$
Thus equation $(2)$ has intger solutions $(X,Y)=(-1,-2).$

Solving for $t^2+t-2=0$, we get $t=1$ and $-2.$
Hence we get integer solution $(x,y)=(1,-2)$ , $(-2,1).$

Answer 3

Well,

let $X=x^2+y^2$ and $Y=x+y$, noting that $2xy=Y^2-X$, so $$XY=2(Y^2-X)+3$$

Solution 1: (a simple one) $$X=\frac{2Y^2+3}{Y+2}$$ $$\implies (Y+2)|(2Y^2+3)$$ but since $Y \equiv-2 \pmod{(Y+2)}$, we get $$2Y^2+3 \equiv11 \equiv0 \pmod{(Y+2)} \implies 11=k(Y+2)$$ Now we can consider $k \in\{\pm1,\pm11\}$ which gives $$Y+2 \in \{11,-11,1,-1\} \implies Y \in \{9,-13,-1,-3\}$$ but since $X>0$, we can't have $Y=-13,-3$ so we can get $Y\in\{9,-1\} \implies X \in\{15,5\}$

and we can easily get systems of equations (in terms of $x+y$ and $xy$) and solve them.

Solution 2: (a more complicated one) $$2Y^2-XY+(3-2X)=0$$ $$\implies Y= \frac{1}{4}\left(X \pm \sqrt{X^2-4(2)(3-2X)}\right)$$ Then, the expression under the root is a perfect square, which gives $$(X+8)^2-88=Z^2$$ We know note that $X >0$ and thus if we let $L=X+8$ we have $L>8$, $$L^2-Z^2=88$$ Now it's easy to consider the cases, and (I actually did it by wolfram alpha) then you can get $$L \in \{23,13\} \implies X \in\{15,5\} \implies Y\in\{9,-1\}$$ and we can also get systems of equations and solve them.

Answer 4

Here I give a different solution. It is clear that x+y is odd as otherwise the left side is even, but the right side is odd.

Also it is clear that $xy \ne 0$.

If $|x+y| \ge 3$, then $|xy| \ge 2$. And $$|(x+y)(x^2+y^2)| \ge 3 \times (x^2+y^2) > 6|xy| > |4xy+3|,$$ which is a contradiction.

Therefore, $|x+y|=1$. If $x+y = 1$, then $0<(x+y)(x^2+y^2)=4xy+3$, so $xy \ge 0$, but there is no solution.

Then we are left with the case $x+y=-1$. We can replace $y$ with $-1-x$ in the original equation and solve it and get x=-2 or 1. Therefore there are exactly two solutions $(x, y) = (-2, 1)$ or $(1,-2).$