Solve for $x,y$ : $x^4+y^4+2 = 4xy$

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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!

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3 Answers

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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,

$$x^4+y^4+2-4xy=(x^2-y^2)^2+2(xy-1)^2$$

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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.

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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 $x=y=±1.$

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