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Question-64204

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Question Number 64204 by naka3546 last updated on 15/Jul/19
Commented by naka3546 last updated on 15/Jul/19
Find all solutions of them if a, b, c ∈ R .
$${Find}\:\:{all}\:\:{solutions}\:\:{of}\:\:{them}\:\:{if}\:\:\:{a},\:{b},\:{c}\:\:\in\:\:\mathbb{R}\:. \\ $$
Answered by MJS last updated on 16/Jul/19
put b=2^n +(1/2^n ) n=0 ⇒ a=(5/2) b=2 c=((17)/4) n=1 ⇒ a=2 b=(5/2) c=−((65)/8) n=2 ⇒ a=−((65)/8) b=((17)/4) c=2 put b=−2^n −(1/2^n ) n=3 ⇒ a=((17)/4) b=−((65)/8) c=(5/2)
$$\mathrm{put}\:{b}=\mathrm{2}^{{n}} +\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${n}=\mathrm{0}\:\Rightarrow\:{a}=\frac{\mathrm{5}}{\mathrm{2}}\:\:{b}=\mathrm{2}\:\:{c}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$${n}=\mathrm{1}\:\Rightarrow\:{a}=\mathrm{2}\:\:{b}=\frac{\mathrm{5}}{\mathrm{2}}\:\:{c}=−\frac{\mathrm{65}}{\mathrm{8}} \\ $$$${n}=\mathrm{2}\:\Rightarrow\:{a}=−\frac{\mathrm{65}}{\mathrm{8}}\:\:{b}=\frac{\mathrm{17}}{\mathrm{4}}\:\:{c}=\mathrm{2} \\ $$$$\mathrm{put}\:{b}=−\mathrm{2}^{{n}} −\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$${n}=\mathrm{3}\:\Rightarrow\:{a}=\frac{\mathrm{17}}{\mathrm{4}}\:\:{b}=−\frac{\mathrm{65}}{\mathrm{8}}\:\:{c}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 16/Jul/19
please elaborate the idea MjS Sir..
$${please}\:{elaborate}\:{the}\:{idea}\:{MjS}\:{Sir}.. \\ $$
Commented by MJS last updated on 16/Jul/19
the idea came at night by discovering the solutions I found by trying different paths had this structure, and looking at the right handed constants 3=((2^2 −1)/2^0 ) ((15)/2)=((2^4 −1)/2^1 ) ((63)/4)=((2^6 −1)/2^2 ) (5/2)=((2^2 +1)/2^1 )=2^1 +(1/2^1 ) ((17)/4)=((2^4 +1)/2^2 )=2^2 +(1/2^2 ) ((65)/8)=((2^6 +1)/2^3 )=2^3 +(1/2^3 ) and 2 can be written as 2^0 +(1/2^0 )
$$\mathrm{the}\:\mathrm{idea}\:\mathrm{came}\:\mathrm{at}\:\mathrm{night}\:\mathrm{by}\:\mathrm{discovering}\:\mathrm{the} \\ $$$$\mathrm{solutions}\:\mathrm{I}\:\mathrm{found}\:\mathrm{by}\:\mathrm{trying}\:\mathrm{different}\:\mathrm{paths} \\ $$$$\mathrm{had}\:\mathrm{this}\:\mathrm{structure},\:\mathrm{and}\:\mathrm{looking}\:\mathrm{at}\:\mathrm{the}\:\mathrm{right} \\ $$$$\mathrm{handed}\:\mathrm{constants} \\ $$$$\mathrm{3}=\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{0}} } \\ $$$$\frac{\mathrm{15}}{\mathrm{2}}=\frac{\mathrm{2}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}^{\mathrm{1}} } \\ $$$$\frac{\mathrm{63}}{\mathrm{4}}=\frac{\mathrm{2}^{\mathrm{6}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }=\mathrm{2}^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} } \\ $$$$\frac{\mathrm{17}}{\mathrm{4}}=\frac{\mathrm{2}^{\mathrm{4}} +\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }=\mathrm{2}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{65}}{\mathrm{8}}=\frac{\mathrm{2}^{\mathrm{6}} +\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }=\mathrm{2}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} } \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\mathrm{2}^{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{0}} } \\ $$
Commented by MJS last updated on 16/Jul/19
but now I see we can substitute a=2cosh α b=2cosh β c=2cosh γ 4(cosh α ∣sinh β∣+cosh β ∣sinh α∣)=3 4(cosh β ∣sinh γ∣+cosh γ ∣sinh β∣)=((15)/2) 4(cosh γ ∣sinh α∣+cosh α ∣sinh γ∣)=((63)/4) α=ln x β=ln y γ=ln z (1/(xy))(∣x^2 −1∣(y^2 +1)+(x^2 +1)∣y^2 −1∣)=3 (1/(yz))(∣y^2 −1∣(z^2 +1)+(y^2 +1)∣z^2 −1∣)=((15)/2) (1/(zx))(∣z^2 −1∣(x^2 +1)+(z^2 +1)∣x^2 −1∣)=((63)/4) ...we get the same 4 solutions as I got before by using all possible combinations of ∣t^2 −1∣=±(t^2 −1) we get 8 solutions but only 4 fit the given equations
$$\mathrm{but}\:\mathrm{now}\:\mathrm{I}\:\mathrm{see}\:\mathrm{we}\:\mathrm{can}\:\mathrm{substitute} \\ $$$${a}=\mathrm{2cosh}\:\alpha \\ $$$${b}=\mathrm{2cosh}\:\beta \\ $$$${c}=\mathrm{2cosh}\:\gamma \\ $$$$\mathrm{4}\left(\mathrm{cosh}\:\alpha\:\mid\mathrm{sinh}\:\beta\mid+\mathrm{cosh}\:\beta\:\mid\mathrm{sinh}\:\alpha\mid\right)=\mathrm{3} \\ $$$$\mathrm{4}\left(\mathrm{cosh}\:\beta\:\mid\mathrm{sinh}\:\gamma\mid+\mathrm{cosh}\:\gamma\:\mid\mathrm{sinh}\:\beta\mid\right)=\frac{\mathrm{15}}{\mathrm{2}} \\ $$$$\mathrm{4}\left(\mathrm{cosh}\:\gamma\:\mid\mathrm{sinh}\:\alpha\mid+\mathrm{cosh}\:\alpha\:\mid\mathrm{sinh}\:\gamma\mid\right)=\frac{\mathrm{63}}{\mathrm{4}} \\ $$$$\alpha=\mathrm{ln}\:{x} \\ $$$$\beta=\mathrm{ln}\:{y} \\ $$$$\gamma=\mathrm{ln}\:{z} \\ $$$$\frac{\mathrm{1}}{{xy}}\left(\mid{x}^{\mathrm{2}} −\mathrm{1}\mid\left({y}^{\mathrm{2}} +\mathrm{1}\right)+\left({x}^{\mathrm{2}} +\mathrm{1}\right)\mid{y}^{\mathrm{2}} −\mathrm{1}\mid\right)=\mathrm{3} \\ $$$$\frac{\mathrm{1}}{{yz}}\left(\mid{y}^{\mathrm{2}} −\mathrm{1}\mid\left({z}^{\mathrm{2}} +\mathrm{1}\right)+\left({y}^{\mathrm{2}} +\mathrm{1}\right)\mid{z}^{\mathrm{2}} −\mathrm{1}\mid\right)=\frac{\mathrm{15}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{zx}}\left(\mid{z}^{\mathrm{2}} −\mathrm{1}\mid\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({z}^{\mathrm{2}} +\mathrm{1}\right)\mid{x}^{\mathrm{2}} −\mathrm{1}\mid\right)=\frac{\mathrm{63}}{\mathrm{4}} \\ $$$$…\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{as}\:\mathrm{I}\:\mathrm{got}\:\mathrm{before} \\ $$$$\mathrm{by}\:\mathrm{using}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{combinations}\:\mathrm{of} \\ $$$$\mid{t}^{\mathrm{2}} −\mathrm{1}\mid=\pm\left({t}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{8}\:\mathrm{solutions}\:\mathrm{but}\:\mathrm{only}\:\mathrm{4}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equations} \\ $$
Commented by mr W last updated on 16/Jul/19
great sir!
$${great}\:{sir}! \\ $$

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