Question Number 9166 by nazar last updated on 21/Nov/16 | ||
$${x}^{\mathrm{831}} +\:{y}^{\mathrm{831}} \:\mathrm{is}\:\mathrm{always}\:\mathrm{divisible}\:\mathrm{by} \\ $$ | ||
Commented by prakash jain last updated on 22/Nov/16 | ||
$${put}\:{x}=−{y} \\ $$$$\left(−{y}\right)^{\mathrm{831}} +{y}^{\mathrm{831}} =\mathrm{0} \\ $$$$\mathrm{Hence}\:{x}+{y}\:\mathrm{is}\:\mathrm{factor}\:\mathrm{of}\:{x}^{\mathrm{831}} +{y}^{\mathrm{831}} \\ $$ | ||
Answered by RasheedSoomro last updated on 22/Nov/16 | ||
$$\mathrm{x}^{\mathrm{831}} +\mathrm{y}^{\mathrm{831}} =\left(\mathrm{x}^{\mathrm{277}} \right)^{\mathrm{3}} +\left(\mathrm{y}^{\mathrm{277}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{x}^{\mathrm{277}} +\mathrm{y}^{\mathrm{277}} \right)\left(\mathrm{x}^{\mathrm{554}} −\mathrm{x}^{\mathrm{277}} \mathrm{y}^{\mathrm{277}} +\mathrm{y}^{\mathrm{554}} \right) \\ $$$$\mathrm{Hence}\:\mathrm{x}^{\mathrm{831}} +\mathrm{y}^{\mathrm{831}} \:\:\mathrm{is}\:\mathrm{always}\:\mathrm{divisible}\:\mathrm{by} \\ $$$$\mathrm{x}^{\mathrm{277}} +\mathrm{y}^{\mathrm{277}} \:,\:\:\mathrm{x}^{\mathrm{554}} −\mathrm{x}^{\mathrm{277}} \mathrm{y}^{\mathrm{277}} +\mathrm{y}^{\mathrm{554}} \:\mathrm{and}\:\:\mathrm{x}^{\mathrm{831}} +\mathrm{y}^{\mathrm{831}} \left(\mathrm{itself}\right) \\ $$ | ||