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Question Number 55902 by naka3546 last updated on 06/Mar/19

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