MHB Real Solutions to x^4+y^4+z^4 = 4xyz-1

[lfdahl]

Find all real solutions to the equation:

$x^4+y^4+z^4 = 4xyz-1$.

 

[Opalg]

Find all real solutions to the equation:

$x^4+y^4+z^4 = 4xyz-1$.

[sp]
Write the equation as $x^4+y^4+z^4 + 1 = 4xyz.$ The left side is positive, so either all three of $x,y,z$ are positive or one of them is positive and the other two are negative. If there is a solution with only one of them positive, then by changing the signs of the other two we get a solution with all three variables positive. Therefore it is enough in the first place to look for solutions where $x,y,z$ are all positive. Then, using the AM-GM inequality, we get $$4xyz = 4\frac{x^4+y^4+z^4 + 1}4 \geqslant 4\sqrt[4]{x^4y^4z^41^4} = 4xyz,$$ with equality only if $x=y=z=1$. Therefore that is the only positive solution. The only other solutions are obtained be changing the signs of two of the variables.

Thus there are altogether four solutions, namely $(x,y,z) = (1,1,1),\ (1,-1,-1),\ (-1,1,-1),\ (-1,-1,1).$

[/sp]

 

[lfdahl]

[sp]
Write the equation as $x^4+y^4+z^4 + 1 = 4xyz.$ The left side is positive, so either all three of $x,y,z$ are positive or one of them is positive and the other two are negative. If there is a solution with only one of them positive, then by changing the signs of the other two we get a solution with all three variables positive. Therefore it is enough in the first place to look for solutions where $x,y,z$ are all positive. Then, using the AM-GM inequality, we get $$4xyz = 4\frac{x^4+y^4+z^4 + 1}4 \geqslant 4\sqrt[4]{x^4y^4z^41^4} = 4xyz,$$ with equality only if $x=y=z=1$. Therefore that is the only positive solution. The only other solutions are obtained be changing the signs of two of the variables.

Thus there are altogether four solutions, namely $(x,y,z) = (1,1,1),\ (1,-1,-1),\ (-1,1,-1),\ (-1,-1,1).$

[/sp]

Thankyou, Opalg, for your participation and for the elegant solution!