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Question Number 22561 by Bruce Lee last updated on 20/Oct/17

Solve for all integer x,y ∈Z       x^2 +y^2 =19

$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{{x}},\boldsymbol{{y}}\:\in\boldsymbol{{Z}} \\ $$$$\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{19} \\ $$

Answered by $@ty@m last updated on 20/Oct/17

Case−1. Let x=0  ⇒y=(√(19))∉Z  Case−2. Let x=±1  ⇒y=(√(18))∉Z  Case−3. Let x=±2  ⇒y=(√(15))∉Z  Case−4. Let x=±3  ⇒y=(√(10))∉Z  Case−5. Let x=±4  ⇒y=(√3)∉Z  ⇒ No solution for x,y∈Z

$${Case}−\mathrm{1}.\:{Let}\:{x}=\mathrm{0} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{19}}\notin{Z} \\ $$$${Case}−\mathrm{2}.\:{Let}\:{x}=\pm\mathrm{1} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{18}}\notin{Z} \\ $$$${Case}−\mathrm{3}.\:{Let}\:{x}=\pm\mathrm{2} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{15}}\notin{Z} \\ $$$${Case}−\mathrm{4}.\:{Let}\:{x}=\pm\mathrm{3} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{10}}\notin{Z} \\ $$$${Case}−\mathrm{5}.\:{Let}\:{x}=\pm\mathrm{4} \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{3}}\notin{Z} \\ $$$$\Rightarrow\:{No}\:{solution}\:{for}\:{x},{y}\in{Z} \\ $$

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