Question Number 31125 by Joel578 last updated on 02/Mar/18
![Let n be a positive integer. Then x^2 + 1 is a factor of (x^4 + 3)^n − [(x^2 + 3)(x^2 − 1)]^n for ... (A) All n (B) Odd n (C) Even n (D) n ≥ 3 (E) None of these options](Q31125.png)
$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}.\:\mathrm{Then}\:{x}^{\mathrm{2}} \:+\:\mathrm{1}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{factor}\:\mathrm{of}\:\left({x}^{\mathrm{4}} \:+\:\mathrm{3}\right)^{{n}} \:−\:\left[\left({x}^{\mathrm{2}} \:+\:\mathrm{3}\right)\left({x}^{\mathrm{2}} \:−\:\mathrm{1}\right)\right]^{{n}} \\ $$$$\mathrm{for}\:... \\ $$$$\left(\mathrm{A}\right)\:\mathrm{All}\:{n} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{Odd}\:{n} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{Even}\:{n} \\ $$$$\left(\mathrm{D}\right)\:{n}\:\geqslant\:\mathrm{3} \\ $$$$\left(\mathrm{E}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{these}\:\mathrm{options} \\ $$
Commented by prof Abdo imad last updated on 04/Mar/18
![if x^2 +1 divide p(x)=(x^4 +3)^(n ) ((x^2 +3)(x^2 −1))^n we must have p(i)=0 and p(−i)=0 but p(i)=4^n −(2(−2))^n =4^n −(−4)^n but we look that p(−i)=p(i)=0 ⇒n≡0[2] ⇒n even.](Q31232.png)
$${if}\:{x}^{\mathrm{2}} \:+\mathrm{1}\:{divide}\:{p}\left({x}\right)=\left({x}^{\mathrm{4}} \:+\mathrm{3}\right)^{{n}\:} \left(\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)\right)^{{n}} \\ $$$${we}\:{must}\:{have}\:{p}\left({i}\right)=\mathrm{0}\:{and}\:{p}\left(−{i}\right)=\mathrm{0}\:{but} \\ $$$${p}\left({i}\right)=\mathrm{4}^{{n}} \:−\left(\mathrm{2}\left(−\mathrm{2}\right)\right)^{{n}} =\mathrm{4}^{{n}} \:−\left(−\mathrm{4}\right)^{{n}} \:{but}\:{we}\:{look} \\ $$$${that}\:{p}\left(−{i}\right)={p}\left({i}\right)=\mathrm{0}\:\Rightarrow{n}\equiv\mathrm{0}\left[\mathrm{2}\right]\:\Rightarrow{n}\:{even}. \\ $$
Answered by supungamage001 last updated on 03/Mar/18
$$\left({c}\right){even} \\ $$$$ \\ $$
Answered by mrW2 last updated on 03/Mar/18
![(x^4 + 3)^n − [(x^2 + 3)(x^2 − 1)]^n =(x^4 +2x^2 +1−2x^2 +2)^n − [(x^2 +1 +2)(x^2 − 1)]^n =[(x^2 +1)^2 −2(x^2 −1)]^n −[(x^2 +1)(x^2 −1)+2(x^2 −1)]^n =(...)+(−2)^n (x^2 −1)^n −(...)−2^n (x^2 −1)^n =(...)+[(−2)^n −2^n ](x^2 −1)^n (...)=terms with factor (x^2 +1) we see that the last term is zero if n is even. ⇒Answer C](Q31169.png)
$$\left({x}^{\mathrm{4}} \:+\:\mathrm{3}\right)^{{n}} \:−\:\left[\left({x}^{\mathrm{2}} \:+\:\mathrm{3}\right)\left({x}^{\mathrm{2}} \:−\:\mathrm{1}\right)\right]^{{n}} \\ $$$$=\left({x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}\right)^{{n}} \:−\:\left[\left({x}^{\mathrm{2}} +\mathrm{1}\:+\mathrm{2}\right)\left({x}^{\mathrm{2}} \:−\:\mathrm{1}\right)\right]^{{n}} \\ $$$$=\left[\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)\right]^{{n}} \:−\left[\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2}\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)\right]^{{n}} \\ $$$$=\left(...\right)+\left(−\mathrm{2}\right)^{{n}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} −\left(...\right)−\mathrm{2}^{{n}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} \\ $$$$=\left(...\right)+\left[\left(−\mathrm{2}\right)^{{n}} −\mathrm{2}^{{n}} \right]\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} \\ $$$$\left(...\right)={terms}\:{with}\:{factor}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${we}\:{see}\:{that}\:{the}\:{last}\:{term}\:{is}\:{zero}\:{if}\:{n}\:{is}\:{even}. \\ $$$$\Rightarrow{Answer}\:{C} \\ $$
Commented by Joel578 last updated on 03/Mar/18
$${thank}\:{you}\:{very}\:{much} \\ $$