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Find all solutions \(y\) (real and otherwise) to the equation \(y^8+16 = 17y^4\)
 

somebody  Oct 19, 2018
 #1
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\(y^8-17y^4+16=0\\ (y^4-16)(y^4 -1) \\ (y^2-4)(y^2+4)(y^2-1)(y^2+1) = \\ (y-2)(y+2)(y+2i)(y-2i)(y+1)(y-1)(y+i)(y-i) \\ \text{and thus roots at }\\ y=\pm 2,~\pm 2i,~\pm 1,~\pm i\)

Rom  Oct 19, 2018

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