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Question Number 118322 by mr W last updated on 16/Oct/20

$$solve$$$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{4}$$$$withx,y,z\in \mathbb{N}$$

Commented by malwaan last updated on 17/Oct/20

$$\frac{yz+xz+xy}{xyz}=\frac{3}{4}(x\ne 0;y\ne 0;z\ne 0)$$$$1)x=y=z$$$$\Rightarrow \frac{3x^2}{x^3}=\frac{3}{4}\Rightarrow \frac{3}{x}=\frac{3}{4}$$$$\Rightarrow x=y=z=4$$$$2)ifx=1\Rightarrow \frac{1}{y}+\frac{1}{z}=-\frac{1}{4}\left\{wrong\right\}$$$$sox>1;y>1;z>1$$$$3)x=2\Rightarrow \frac{1}{y}+\frac{1}{z}=\frac{1}{4}$$$$\Rightarrow y=8;z=8$$$$4)x=3\Rightarrow \frac{1}{y}+\frac{1}{z}=\frac{5}{12}$$$$\frac{y+z}{yz}=\frac{5}{12}\left\{nosolutions\right\}$$$$sox\ne 3;y\ne 3;z\ne 3$$

Commented by MJS_new last updated on 17/Oct/20

$$\mathrm{with}x⩽y⩽z\mathrm{I}\mathrm{found}\mathrm{these}\mathrm{triples}$$$$2520$$$$2612$$$$288$$$$3312$$$$346$$$$444$$

Answered by TANMAY PANACEA last updated on 16/Oct/20

$$x=y=z=4$$$$x=2y=z=8$$$$itisnotsolution..bytrial$$

Answered by 1549442205PVT last updated on 17/Oct/20

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\left(1\right)$$$$\mathrm{Since}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{have}\mathrm{equal}\mathrm{role}\mathrm{in}\mathrm{the}\mathrm{equality}$$$$,\mathrm{WLOG}\mathrm{we}\mathrm{can}\mathrm{assume}\mathrm{that}\mathrm{x}⩾\mathrm{y}⩾\mathrm{z}>1$$$$\mathrm{Then}\frac{1}{\mathrm{x}}⩽\frac{1}{\mathrm{y}}⩽\frac{1}{\mathrm{z}}<1.\mathrm{Hence}$$$$\frac{3}{4}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}⩽\frac{3}{\mathrm{z}}\Rightarrow \mathrm{z}⩽4$$$$\mathrm{i})\mathrm{For}\mathrm{z}=4\mathrm{substituting}\mathrm{into}\left(1\right)\mathrm{we}\mathrm{get}$$$$\frac{2}{\mathrm{y}}⩾\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{1}{2}\left(2\right)\Rightarrow \mathrm{y}⩽4\Rightarrow \mathrm{y}\in \{1,2,3,4\}$$$$\bullet \mathrm{For}\mathrm{y}=4\mathrm{substituting}\mathrm{into}\left(2\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\Rightarrow \mathrm{x}=4$$$$\bullet \mathrm{For}\mathrm{y}=3\mathrm{substituting}\mathrm{into}\left(2\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\Rightarrow \mathrm{x}=6$$$$\bullet \mathrm{For}\mathrm{y}\in \{1,2\}\mathrm{substituting}\mathrm{into}\left(2\right)\mathrm{we}$$$$\mathrm{get}\mathrm{L}.\mathrm{H}.\mathrm{S}\left(2\right)>\frac{1}{2}\Rightarrow \mathrm{has}\mathrm{no}\mathrm{roots}$$$$\mathrm{ii})\mathrm{For}\mathrm{z}=3\mathrm{substituting}\mathrm{into}\left(1\right)\mathrm{we}\mathrm{get}$$$$\frac{2}{\mathrm{y}}⩾\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\left(3\right)\Rightarrow 5\mathrm{y}⩽24.$$$$\mathrm{Since}\mathrm{y}\in \mathrm{N}\Rightarrow \mathrm{y}\in \{1,2,3,4\}$$$$\bullet \mathrm{For}\mathrm{y}=4\mathrm{substituting}\mathrm{into}\left(3\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{5}{12}-\frac{1}{4}=\frac{2}{12}=\frac{1}{6}\Rightarrow \mathrm{x}=6$$$$\bullet \mathrm{For}\mathrm{y}=3\mathrm{substituting}\mathrm{into}\left(3\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{5}{12}-\frac{1}{3}=\frac{1}{12}\Rightarrow \mathrm{x}=12$$$$\bullet \mathrm{For}\mathrm{y}\in \{1,2\}\mathrm{substituting}\mathrm{into}\left(3\right)\mathrm{we}$$$$\mathrm{get}\mathrm{L}.\mathrm{H}.\mathrm{S}\left(3\right)>\frac{5}{12}\Rightarrow \mathrm{has}\mathrm{no}\mathrm{roots}$$$$\mathrm{iii})\mathrm{For}\mathrm{z}=2\mathrm{substituting}\mathrm{into}\left(1\right)\mathrm{we}\mathrm{get}$$$$\frac{2}{\mathrm{y}}⩾\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\left(4\right)\Rightarrow \mathrm{y}⩽8$$$$\bullet \mathrm{For}\mathrm{y}=8\mathrm{substituting}\mathrm{into}\left(4\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}\Rightarrow \mathrm{x}=8$$$$\bullet \mathrm{For}\mathrm{y}=7\mathrm{substituting}\mathrm{into}\left(4\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{1}{4}-\frac{1}{7}=\frac{3}{28}\Rightarrow 3\mathrm{x}=28\Rightarrow \mathrm{x}\notin \mathrm{N}$$$$\bullet \mathrm{For}\mathrm{y}=6\mathrm{substituting}\mathrm{into}\left(4\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\Rightarrow \mathrm{x}=12$$$$\bullet \mathrm{For}\mathrm{y}=5\mathrm{substituting}\mathrm{into}\left(4\right)\mathrm{we}\mathrm{get}$$$$\frac{1}{\mathrm{x}}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\Rightarrow \mathrm{x}=20$$$$\bullet \mathrm{For}\mathrm{y}\in \{1,2,3,4\}\mathrm{substituting}\mathrm{into}\left(4\right)$$$$\mathrm{we}\mathrm{get}\mathrm{L}.\mathrm{H}.\mathrm{S}\left(4\right)>\frac{1}{4}\Rightarrow \mathrm{has}\mathrm{no}\mathrm{roots}$$$$\mathrm{Combining}\mathrm{three}\mathrm{above}\mathrm{cases}\mathrm{we}\mathrm{obtain}$$$$6\mathrm{roots}\mathrm{are}:$$$$(\mathrm{x},\mathrm{y},\mathrm{z})\in \{(4,4,4),(6,4,3),(12,3,3)$$$$(8,8,2),(12,6,2),(20,5,2)\}(\ast )$$$$\mathbf{Consequently},\mathbf{the}\mathbf{equation}$$$$\frac{1}{\mathit{x}}+\frac{1}{\mathit{y}}+\frac{1}{\mathit{z}}=\frac{3}{4}\mathbf{has}3.3!+2.\frac{3!}{2}+1=25$$$$\mathbf{roots}(\mathbf{x},\mathbf{y},\mathbf{z}).\mathbf{They}\mathbf{are}\mathbf{elements}\mathbf{of}(\ast )$$$$\mathbf{and}\mathbf{all}\mathbf{their}\mathbf{permutation}$$

Commented by mr W last updated on 17/Oct/20

$$thankyouallsirs!$$

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