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Question Number 47310 by Meritguide1234 last updated on 08/Nov/18

Commented by MJS last updated on 08/Nov/18

x≠y≠z≠0 I guess.  but we know that there′s no triple xyz for n>2  so we only have to show for n=2 and in this  case it′s obviously true

$${x}\neq{y}\neq{z}\neq\mathrm{0}\:\mathrm{I}\:\mathrm{guess}. \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{triple}\:{xyz}\:\mathrm{for}\:{n}>\mathrm{2} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{only}\:\mathrm{have}\:\mathrm{to}\:\mathrm{show}\:\mathrm{for}\:{n}=\mathrm{2}\:\mathrm{and}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{case}\:\mathrm{it}'\mathrm{s}\:\mathrm{obviously}\:\mathrm{true}\: \\ $$

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