The number of triples (x y z) of real numbers satisfying the equation x^4+y^4+z^4+1=4xyz
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The left side is equal to (x2−y2)2+2x2y2+z4−4xyz(x2−y2)2+2x2y2+z4−4xyz. The presence of 2x2y22x2y2 and −4xyz−4xyz suggests the possibility of adding a 2z22z2 and then completing one more square. So the equation can be rewritten as (x2−y2)2+(z2−1)2+2(xy−z)2=0(x2−y2)2+(z2−1)2+2(xy−z)2=0. This equality can hold only if all three squares are equal to zero. From z2−1=0z2−1=0 we have z=±1z=±1, and after a quick analysis we conclude that the solutions are (1,1,1),(−1,1,−1),(1,−1,−1),(−1,−1,1)(1,1,1),(−1,1,−1),(1,−1,−1),(−1,−1,1).
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The left side is equal to (x2−y2)2+2x2y2+z4−4xyz(x2−y2)2+2x2y2+z4−4xyz. The presence of 2x2y22x2y2 and −4xyz−4xyz suggests the possibility of adding a 2z22z2 and then completing one more square. So the equation can be rewritten as (x2−y2)2+(z2−1)2+2(xy−z)2=0(x2−y2)2+(z2−1)2+2(xy−z)2=0. This equality can hold only if all three squares are equal to zero. From z2−1=0z2−1=0 we have z=±1z=±1, and after a quick analysis we conclude that the solutions are (1,1,1),(−1,1,−1),(1,−1,−1),(−1,−1,1)(1,1,1),(−1,1,−1),(1,−1,−1),(−1,−1,1).
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