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Question Number 152481 by mathdanisur last updated on 28/Aug/21

Commented by Rasheed.Sindhi last updated on 29/Aug/21

$S o m e s o l u t i o n s :$ $(x, y, z) = (3, 3, 8), (3, 8, 3), (8, 3, 3)$
	Answered by Rasheed.Sindhi last updated on 29/Aug/21
	$(1+(1/x))(1+(1/y))(1+(1/z))=2; x,y,z∈Z^+ (x+1)(y+1)(z+1)=2xyz ⇒At least one of x+1,y+1,z+1 is even ⇒At least one of x,y,z is an odd. Let x=2k−1: 2k(y+1)(z+1)=2(2k−1)yz k(y+1)(z+1)=(2k−1)yz kyz+ky+kz+k=(2k−1)yz (2k−1)yz−kyz=ky+kz+k (k−1)yz=k(y+z+1) ⇒k∣(k−1)yz ⇒ { ((k∣(k−1)⇒k=1)),((k∣yz⇒k∣yz ∨ k∣y ∨ k∣z)) :} Case1: k=1⇒x=2k−1=2(1)−1=1 (1+(1/x))(1+(1/y))(1+(1/z))=2 ⇒(1+(1/1))(1+(1/y))(1+(1/z))=2 (1+(1/y))(1+(1/z))=1 1+(1/y)=(1+(1/z))^(−1) =(z/(z+1)) ((y+1)/y)=(z/(z+1)) yz=yz+y+z+1⇒y+z=−1(×) (∵y,z∈Z^+ ⇒y+z ≠−1) k≠1⇒x≠1⇒y≠1⇒z≠1 Case2: k∣y (k−1)yz=k(y+z+1) y=km: km(k−1)z=k(km+z+1) m(k−1)z=km+z+1 m(k−1)z−z=km+1 (km−m−1)z=km+1 z=((km+1)/(km−m−1)) k=2,m=4⇒ z=((8+1)/(8−4−1))=(9/3)=3 y=8 ⇒x=2(2)−1=3 Verification: (1+(1/x))(1+(1/y))(1+(1/z))=2; x,y,z∈Z^+ (1+(1/3))(1+(1/8))(1+(1/3))=2 (4/3)×(9/8)×(4/3)=2 2=2 Continue...$
	$(1 + \frac{1}{x}) (1 + \frac{1}{y}) (1 + \frac{1}{z}) = 2; x, y, z \in Z^{+}$ $(x + 1) (y + 1) (z + 1) = 2 x y z$ $\Rightarrow A t l e a s t o n e o f x + 1, y + 1, z + 1 i s e v e n$ $\Rightarrow A t l e a s t o n e o f x, y, z i s a n o d d .$ $L e t x = 2 k - 1 :$ $2 k (y + 1) (z + 1) = 2 (2 k - 1) y z$ $k (y + 1) (z + 1) = (2 k - 1) y z$ $k y z + k y + k z + k = (2 k - 1) y z$ $(2 k - 1) y z - k y z = k y + k z + k$ $(k - 1) y z = k (y + z + 1)$ $\Rightarrow k ∣ (k - 1) y z$ $\Rightarrow {\begin{cases} k ∣ (k - 1) \Rightarrow k = 1 \\ k ∣ y z \Rightarrow k ∣ y z \lor k ∣ y \lor k ∣ z \end{cases}$ $C a s e 1 : k = 1 \Rightarrow x = 2 k - 1 = 2 (1) - 1 = 1$ $(1 + \frac{1}{x}) (1 + \frac{1}{y}) (1 + \frac{1}{z}) = 2$ $\Rightarrow (1 + \frac{1}{1}) (1 + \frac{1}{y}) (1 + \frac{1}{z}) = 2$ $(1 + \frac{1}{y}) (1 + \frac{1}{z}) = 1$ $1 + \frac{1}{y} = {(1 + \frac{1}{z})}^{- 1} = \frac{z}{z + 1}$ $\frac{y + 1}{y} = \frac{z}{z + 1}$ $y z = y z + y + z + 1 \Rightarrow y + z = - 1 (\times)$ $(∵ y, z \in Z^{+} \Rightarrow y + z \neq - 1)$ $k \neq 1 \Rightarrow x \neq 1 \Rightarrow y \neq 1 \Rightarrow z \neq 1$ $Case 2 : \underset{―}{} k ∣ y \underset{―}{}$ $(k - 1) y z = k (y + z + 1)$ $y = k m : k m (k - 1) z = k (k m + z + 1)$ $m (k - 1) z = k m + z + 1$ $m (k - 1) z - z = k m + 1$ $(k m - m - 1) z = k m + 1$ $z = \frac{k m + 1}{k m - m - 1}$ $k = 2, m = 4 \Rightarrow$ $z = \frac{8 + 1}{8 - 4 - 1} = \frac{9}{3} = 3$ $y = 8 \Rightarrow x = 2 (2) - 1 = 3$ $V e r i f i c a t i o n :$ $(1 + \frac{1}{x}) (1 + \frac{1}{y}) (1 + \frac{1}{z}) = 2; x, y, z \in Z^{+}$ $(1 + \frac{1}{3}) (1 + \frac{1}{8}) (1 + \frac{1}{3}) = 2$ $\frac{4}{3} \times \frac{9}{8} \times \frac{4}{3} = 2$ $2 = 2$ $C o n t i n u e . . .$
	Commented by mathdanisur last updated on 29/Aug/21

	$Very nice Ser$ $Ans : (3; 8; 3) (. . . .) (3; 4; 5) (. . .) (2; 5; 9) (. . .) (2; 6; 7) (. . .) (2; 4; 15) (. . .)$
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