The equation $x^4+y^4+z^4 = 4xyz-1$ has four real solutions: $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, and $(-1,-1,1)$. The analysis begins by rewriting the equation as $x^4+y^4+z^4 + 1 = 4xyz$, establishing that the left side is positive. By applying the AM-GM inequality, it is determined that the only positive solution occurs when $x=y=z=1$. Other solutions arise from changing the signs of two variables, confirming the total of four distinct solutions.
PREREQUISITESMathematicians, students studying algebra, and anyone interested in solving polynomial equations and understanding inequalities.
[sp]lfdahl said:Find all real solutions to the equation:
$x^4+y^4+z^4 = 4xyz-1$.
Opalg said:[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 $(x,y,z) = (1,1,1),\ (1,-1,-1),\ (-1,1,-1),\ (-1,-1,1).$
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