anemone said:
Find all real solutions of the equation $$8x^2-11y^2-40x-8y+4xy+4y^2x+y^4+52=0$$
My solution:
First, I group $$4xy+4y^2x$$ and factor to get:
$$4xy+4y^2x=4xy(1+y)$$
I notice that $$(1+y)$$ is a factor of $$4xy+4y^2x$$ and my plan is to rewrite the given expression as the sum of two terms and next equate them to zero.
To do this, I need to group a certain amount of terms and hopefully, there will be the factor of $$(1+y)$$ in them so that I could further simplify the expression.
For $$y^4$$ in the remaining terms of the given expression $$8x^2-11y^2-40x-8y+\cancel {4xy+4y^2x}+y^4+52=0$$:
I add a $$y^3$$ in the equation and but I will subtract it next to maintain the equality.
$$y^4+y^3=y^3(y+1)$$
For $$-11y^2-8y-y^3$$ in $$8x^2-11y^2-40x-8y+\cancel {4xy+4y^2x}+\cancel {y^4}+52-y^3=0$$:
I know that if I want to let $$(1+y)$$ be a factor of this cubic equation, I must add a constant of 2 to the equation and therefore I will proceed like this:
$$-11y^2-8y-y^3+2=-\left(y^3+11y^2+8y-2 \right)=-(1+y)\left(y^2+10y-2 \right)$$
Now, for the remaining terms in $$8x^2-\cancel {11y^2}-40x-\cancel {8y}+\cancel {4xy+4y^2x}+\cancel {y^4}+50-\cancel {y^3}=0$$, I see that I could rewrite it to become:
$$8x^2-40x+50=2(2x-5)^2$$
Putting all these pieces together yields:
$$4xy(1+y)+y^3(y+1)-(1+y)\left(y^2+10y-2 \right)+2(2x-5)^2=0$$
$$(1+y)\left(4xy+y^3-\left(y^2+10y-2 \right) \right)+2(2x-5)^2=0$$
$$(1+y)\left(4xy+y^3-y^2-10y+2 \right)+2(2x-5)^2=0$$
Clearly $$x=\dfrac{5}{2}$$ and $$y=-1$$ is one of the valid answers to the problem.
I suspect the given equation can be rewritten as the sum of two terms in different way as well so let's see:
From $$-8y+4xy+4y^2x$$, I simplify it to get:
$$-8y+4xy+4y^2x=4y(-2+x+xy)$$ (*)
We know from a well-known identity that $$xy+x-y-1=(x-1)(y+1)$$
Thus, (*) can be manipulated a bit to obtain the following:
$$-8y+4xy+4y^2x=4y(-1-y+x+xy)+4y^2-4y=4y(x-1)(y+1)+4y^2-4y$$
Now, we know we must rewrite the rest of the given terms with $$(x-1)$$ as one of the factors. If we focus on the terms $$8x^2-40x$$ from $$8x^2-11y^2-40x-\cancel {8y+4xy+4y^2x}+y^4+52=0$$, we see that we may rewrite it by having the factor $$(x-1)$$ iff we have a constant of 32 in the equation, like this:
$$8x^2-40x+32=8(x-1)(x-4)$$
Next, by replacing these two newly rearranged terms within the equation we find that now we have:
$$4y(x-1)(y+1)+4y^2-4y+8(x-1)(x-4)+y^4-11y^2+32=0$$
or
$$(x-1)(4y(y+1)+8(x-4))+4y^2-4y+y^4-11y^2+32=0$$
$$(x-1)(4y(y+1)+8(x-4))+y^4-7y^2-4y+32=0$$
$$(x-1)(4y(y+1)+8(x-4))+(y-2)^2(y^2+4y+5)=0$$
We can see that $$x=1$$ and $$y=2$$ is another pair of the valid answers to the problem.
I find that there are no other ways to regroup the terms and rewrite it as the sum of two terms and I conclude that the solutions (real) to the problem are
$$(x,y)=\left(\frac{5}{2},-1 \right),\,(1,2)$$.
