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Question-55902

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Question Number 55902 by naka3546 last updated on 06/Mar/19
Answered by MJS last updated on 06/Mar/19
P_n =(a+(1/2))^n +(a+1+(1/2))^n ; a∈Z^+ P_1 =2a+2 ∈Z P_2 =2a^2 +4a+(5/2) ∉Z P_3 =2a^3 +6a^2 +((15)/2)a+(7/2) ∈Z∣a=2k−1; k∈Z P_4 =...+((41)/8) ∉Z P_5 =...+((205)/8)a+((61)/8) ∈Z∣a=8k−1; k∈Z P_(2m) ∉Z; m∈Z P_7 ∈Z∣a=32k−1; k∈Z P_9 ∈Z∣a=128k−1; k∈Z P_(2m−1) ∈Z∣a=2^(2m−3) k−1; k, m∈Z ⇒ solve for k, m∈Z: 2018=2^(2m−3) k−1 k(m)=((2019)/2^(2m−3) )=((16152)/4^m ) ⇒ m≤1 n=2m−1>0 ⇒ m=1 ⇒ n=1 ⇒ only P_1 is integer testing for P_n =(2019+(1/2))^n +(2020+(1/2))^n k(m)=((2020)/2^(2m−3) )=((16160)/4^m ) ⇒ m≤2 ⇒ n=1 ∨ n=3 P_1 and P_3 are integer btw. for a=2047 P_i are integer for i∈{1, 3, 5, 7, 9, 11, 13}
$${P}_{{n}} =\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +\left({a}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} ;\:{a}\in\mathbb{Z}^{+} \\ $$$${P}_{\mathrm{1}} =\mathrm{2}{a}+\mathrm{2}\:\in\mathbb{Z} \\ $$$${P}_{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} +\mathrm{4}{a}+\frac{\mathrm{5}}{\mathrm{2}}\:\notin\mathbb{Z} \\ $$$${P}_{\mathrm{3}} =\mathrm{2}{a}^{\mathrm{3}} +\mathrm{6}{a}^{\mathrm{2}} +\frac{\mathrm{15}}{\mathrm{2}}{a}+\frac{\mathrm{7}}{\mathrm{2}}\:\in\mathbb{Z}\mid{a}=\mathrm{2}{k}−\mathrm{1};\:{k}\in\mathbb{Z} \\ $$$${P}_{\mathrm{4}} =…+\frac{\mathrm{41}}{\mathrm{8}}\:\notin\mathbb{Z} \\ $$$${P}_{\mathrm{5}} =…+\frac{\mathrm{205}}{\mathrm{8}}{a}+\frac{\mathrm{61}}{\mathrm{8}}\:\in\mathbb{Z}\mid{a}=\mathrm{8}{k}−\mathrm{1};\:{k}\in\mathbb{Z} \\ $$$${P}_{\mathrm{2}{m}} \:\notin\mathbb{Z};\:{m}\in\mathbb{Z} \\ $$$${P}_{\mathrm{7}} \:\in\mathbb{Z}\mid{a}=\mathrm{32}{k}−\mathrm{1};\:{k}\in\mathbb{Z} \\ $$$${P}_{\mathrm{9}} \:\in\mathbb{Z}\mid{a}=\mathrm{128}{k}−\mathrm{1};\:{k}\in\mathbb{Z} \\ $$$${P}_{\mathrm{2}{m}−\mathrm{1}} \:\in\mathbb{Z}\mid{a}=\mathrm{2}^{\mathrm{2}{m}−\mathrm{3}} {k}−\mathrm{1};\:{k},\:{m}\in\mathbb{Z} \\ $$$$\Rightarrow\:\mathrm{solve}\:\mathrm{for}\:{k},\:{m}\in\mathbb{Z}:\:\mathrm{2018}=\mathrm{2}^{\mathrm{2}{m}−\mathrm{3}} {k}−\mathrm{1} \\ $$$${k}\left({m}\right)=\frac{\mathrm{2019}}{\mathrm{2}^{\mathrm{2}{m}−\mathrm{3}} }=\frac{\mathrm{16152}}{\mathrm{4}^{{m}} } \\ $$$$\Rightarrow\:{m}\leqslant\mathrm{1} \\ $$$${n}=\mathrm{2}{m}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:{m}=\mathrm{1}\:\Rightarrow\:{n}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{only}\:{P}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{integer} \\ $$$$ \\ $$$$\mathrm{testing}\:\mathrm{for}\:{P}_{{n}} =\left(\mathrm{2019}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +\left(\mathrm{2020}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${k}\left({m}\right)=\frac{\mathrm{2020}}{\mathrm{2}^{\mathrm{2}{m}−\mathrm{3}} }=\frac{\mathrm{16160}}{\mathrm{4}^{{m}} } \\ $$$$\Rightarrow\:{m}\leqslant\mathrm{2}\:\Rightarrow\:{n}=\mathrm{1}\:\vee\:{n}=\mathrm{3} \\ $$$${P}_{\mathrm{1}} \:\mathrm{and}\:{P}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{integer} \\ $$$$\mathrm{btw}.\:\mathrm{for}\:{a}=\mathrm{2047}\:{P}_{{i}} \:\mathrm{are}\:\mathrm{integer}\:\mathrm{for}\:{i}\in\left\{\mathrm{1},\:\mathrm{3},\:\mathrm{5},\:\mathrm{7},\:\mathrm{9},\:\mathrm{11},\:\mathrm{13}\right\} \\ $$

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