Let-n-be-an-even-positive-integer-such-that-n-2-is-odd-and-let-0-1-n-1-be-the-complex-roots-of-unity-of-order-n-Prove-that-k-0-n-1-a-b-k-2-a-n-2-b-n-2-2

Question Number 21293 by Tinkutara last updated on 19/Sep/17

$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{an}\:\mathrm{even}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{such} \\ $$$$\mathrm{that}\:\frac{{n}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{let}\:\alpha_{\mathrm{0}} ,\:\alpha_{\mathrm{1}} ,\:….,\:\alpha_{{n}−\mathrm{1}} \:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{of}\:\mathrm{order}\:{n}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}\:+\:{b}\alpha_{{k}} ^{\mathrm{2}} \right)\:=\:\left({a}^{\frac{{n}}{\mathrm{2}}} \:+\:{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{any}\:\mathrm{complex}\:\mathrm{numbers}\:{a}\:\mathrm{and}\:{b}. \\ $$

Answered by revenge last updated on 24/Sep/17

$${x}^{{n}} −\mathrm{1}=\left({x}−\alpha_{\mathrm{0}} \right)\left({x}−\alpha_{\mathrm{1}} \right)\left({x}−\alpha_{\mathrm{2}} \right)…\left({x}−\alpha_{{n}−\mathrm{1}} \right) \\ $$$${Putting}\:{x}={i}\sqrt{\frac{{a}}{{b}}},\:{we}\:{get} \\ $$$${i}^{{n}} \left(\frac{{a}}{{b}}\right)^{\frac{{n}}{\mathrm{2}}} −\mathrm{1}=\left({i}\sqrt{\frac{{a}}{{b}}}−\alpha_{\mathrm{0}} \right)\left({i}\sqrt{\frac{{a}}{{b}}}−\alpha_{\mathrm{1}} \right)…\left({i}\sqrt{\frac{{a}}{{b}}}−\alpha_{{n}−\mathrm{1}} \right) \\ $$$${i}^{{n}} {a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} =\left({i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{0}} \right)\left({i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{1}} \right)…\left({i}\sqrt{{a}}−\sqrt{{b}}\alpha_{{n}−\mathrm{1}} \right)\:…\left(\mathrm{1}\right) \\ $$$${Similarly}\:{putting}\:{x}=−{i}\sqrt{\frac{{a}}{{b}}},\:{we}\:{get} \\ $$$$\left(−{i}^{{n}} \right){a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} =\left(−{i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{0}} \right)\left(−{i}\sqrt{{a}}−\sqrt{{b}}\alpha_{\mathrm{1}} \right)…\left(−{i}\sqrt{{a}}−\sqrt{{b}}\alpha_{{n}−\mathrm{1}} \right) \\ $$$${i}^{{n}} {a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} =\left({i}\sqrt{{a}}+\sqrt{{b}}\alpha_{\mathrm{0}} \right)\left({i}\sqrt{{a}}+\sqrt{{b}}\alpha_{\mathrm{1}} \right)…\left({i}\sqrt{{a}}+\sqrt{{b}}\alpha_{{n}−\mathrm{1}} \right)\:…\left(\mathrm{2}\right) \\ $$$${Multiplying}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right),\:{we}\:{get} \\ $$$$\left({i}^{{n}} {a}^{\frac{{n}}{\mathrm{2}}} −{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} =\left({a}+{b}\alpha_{\mathrm{0}} ^{\mathrm{2}} \right)\left({a}+{b}\alpha_{\mathrm{1}} ^{\mathrm{2}} \right)…\left({a}+{b}\alpha_{{n}−\mathrm{1}} ^{\mathrm{2}} \right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}+{b}\alpha_{{k}} ^{\mathrm{2}} \right) \\ $$$${Since}\:{n}=\mathrm{2}{m},\:{where}\:{m}\:{is}\:{odd},\:\therefore\:{i}^{\mathrm{2}{m}} =−\mathrm{1}.\:{So}, \\ $$$$\left({a}^{\frac{{n}}{\mathrm{2}}} +{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}+{b}\alpha_{{k}} ^{\mathrm{2}} \right) \\ $$

Commented by Tinkutara last updated on 24/Sep/17

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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