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Question Number 9166 by nazar last updated on 21/Nov/16

x^(831) + y^(831)  is always divisible by

$${x}^{\mathrm{831}} +\:{y}^{\mathrm{831}} \:\mathrm{is}\:\mathrm{always}\:\mathrm{divisible}\:\mathrm{by} \\ $$

Commented by prakash jain last updated on 22/Nov/16

put x=−y  (−y)^(831) +y^(831) =0  Hence x+y is factor of x^(831) +y^(831)

$${put}\:{x}=−{y} \\ $$$$\left(−{y}\right)^{\mathrm{831}} +{y}^{\mathrm{831}} =\mathrm{0} \\ $$$$\mathrm{Hence}\:{x}+{y}\:\mathrm{is}\:\mathrm{factor}\:\mathrm{of}\:{x}^{\mathrm{831}} +{y}^{\mathrm{831}} \\ $$

Answered by RasheedSoomro last updated on 22/Nov/16

x^(831) +y^(831) =(x^(277) )^3 +(y^(277) )^3                        =(x^(277) +y^(277) )(x^(554) −x^(277) y^(277) +y^(554) )  Hence x^(831) +y^(831)   is always divisible by  x^(277) +y^(277)  ,  x^(554) −x^(277) y^(277) +y^(554)  and  x^(831) +y^(831) (itself)

$$\mathrm{x}^{\mathrm{831}} +\mathrm{y}^{\mathrm{831}} =\left(\mathrm{x}^{\mathrm{277}} \right)^{\mathrm{3}} +\left(\mathrm{y}^{\mathrm{277}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{x}^{\mathrm{277}} +\mathrm{y}^{\mathrm{277}} \right)\left(\mathrm{x}^{\mathrm{554}} −\mathrm{x}^{\mathrm{277}} \mathrm{y}^{\mathrm{277}} +\mathrm{y}^{\mathrm{554}} \right) \\ $$$$\mathrm{Hence}\:\mathrm{x}^{\mathrm{831}} +\mathrm{y}^{\mathrm{831}} \:\:\mathrm{is}\:\mathrm{always}\:\mathrm{divisible}\:\mathrm{by} \\ $$$$\mathrm{x}^{\mathrm{277}} +\mathrm{y}^{\mathrm{277}} \:,\:\:\mathrm{x}^{\mathrm{554}} −\mathrm{x}^{\mathrm{277}} \mathrm{y}^{\mathrm{277}} +\mathrm{y}^{\mathrm{554}} \:\mathrm{and}\:\:\mathrm{x}^{\mathrm{831}} +\mathrm{y}^{\mathrm{831}} \left(\mathrm{itself}\right) \\ $$

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