Question-185269

Question Number 185269 by Rupesh123 last updated on 19/Jan/23

Answered by aleks041103 last updated on 19/Jan/23

p = \sqrt{n - 8} + \sqrt{n + 8} \Rightarrow n ⩾ 8

p^{2} = 2 n - 2 \sqrt{n^{2} - 64}

\Rightarrow \sqrt{4 (n^{2} - 64)} = 2 n - p^{2}

s i n c e n, p \in Z \Rightarrow \sqrt{4 (n^{2} - 64)} \in Z

\Rightarrow \exists m \in Z \Rightarrow 4 n^{2} - 16^{2} = m^{2}

\Rightarrow 4 ∣ m^{2} \Rightarrow 2 ∣ m, s o l e t m \to 2 m

\Rightarrow 4 n^{2} - 16^{2} = 4 m^{2}

\Rightarrow n^{2} - 8^{2} = m^{2} \Rightarrow (n - m) (n + m) = 8^{2} = 2^{6}

a l s o n ⩾ 8 \Rightarrow n + m ⩾ 8

\Rightarrow n + m = 8, 16, 32, 64

n - m = 8, 4, 2, 1

s o

(n, m) \in {(8, 0), (10, 6), (17, 15)}

t h i s i s n e c e s s a r y b u t i n s u f f i c i e n t

a f t e f c h e c k

n = 8 \Rightarrow p = \sqrt{8 - 8} + \sqrt{8 + 8} = 4 \in Z

n = 10 \Rightarrow p = \sqrt{10 - 8} + \sqrt{10 + 8} = 10 \sqrt{2} \notin Z

n = 17 \Rightarrow p = \sqrt{17 - 8} + \sqrt{17 + 8} = 8 \in Z

\Rightarrow n \in {8, 17}

A n s . 25

Answered by Frix last updated on 19/Jan/23

u = \sqrt{n - 8} \Rightarrow n = u^{2} + 8 \Rightarrow

\sqrt{n + 8} = \sqrt{u^{2} + 16} = v \in Z \Rightarrow

u^{2} + 16 = v^{2}

v^{2} - u^{2} = 16

v = u + w

(2 u + w) w = 16

\Rightarrow

w = 1 \land 2 u + 1 = 16 impossible

w = 2 \land 2 u + 2 = 8 \Rightarrow u = 3 \Rightarrow n = 17 ★

w = 4 \land 2 u + 4 = 4 \Rightarrow u = 0 \Rightarrow n = 8 ★

w = 8 \land 2 u + 8 = 2 impossible

w = 16 \land 2 u + 16 = 1 impossible

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