Solve for $x,y$ : $x^4+y^4+2 = 4xy$
Obviously, the very obvious solutions are $(1,1)$ and $(-1,-1)$.
But I don't know how to reach the answer. I got the answer through hit and trial.
I tried adding and subtracting $2x^2y^2$ in the $LHS$:
$x^4+y^4+2 +2x^2y^2-2x^2y^2 = 4xy$
$(x^4+y^4)^2 = 4xy + 2x^2y^2 -2$
Giving us:
$(x^4+y^4)^2 =2(2xy + x^2y^2 -1)$
How to proceed further? I don't really think this helps. Is there a better method? Any hints or solutions would be appreciated!
Hint:
Using AM-GM inequality for real $x^2,y^2\ge0$
$$\dfrac{x^4+y^4+1+1}4\ge\sqrt[4]{x^4y^4}$$
the equality occurs if $x^4=y^4=1$
Alternatively,
$$x^4+y^4+2-4xy=(x^2-y^2)^2+2(xy-1)^2$$
From your equation $x^4+y^4+2x^2y^2-2x^2y^2+2=4xy$ it follows $$x^4+y^4-2x^2y^2=4xy-2-2x^2y^2 \Longleftrightarrow (x^2-y^2)^2=-2(xy-1)^2$$
Over the real numbers this is possible if and only if $$x^2=y^2 \text{ and } xy=1$$ which has only $(1,1)$ and $(-1,-1)$ as solutions.
After realizing that $x^2, y^2≥0$, then
$$x^4+1≥2x^2$$
$$y^4+1≥2y^2$$
$$x^4+y^4+2≥2(x^2+y^2)$$
$$x^4+y^4+2-4xy≥2(x^2+y^2)-4xy=2(x-y)^2≥0$$
So, the solution occurs for only $x=y$, which follows $x=y=±1.$
