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Question Number 212926 by efronzo1 last updated on 27/Oct/24
find integers x,y such that (x/(x−3)) −(4/(y^2 −45)) = (1/(100))
$$\:\mathrm{find}\:\mathrm{integers}\:\mathrm{x},\mathrm{y}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{3}}\:−\frac{\mathrm{4}}{\mathrm{y}^{\mathrm{2}} −\mathrm{45}}\:=\:\frac{\mathrm{1}}{\mathrm{100}} \\ $$
Answered by Frix last updated on 27/Oct/24
⇔ x=((3(y^2 +355))/(4855−99y^2 )) x∈Z ⇒ 3∣y^2 +355∣≥∣4855−99y^2 ∣ ⇒ 3(y^2 +355)≥∣4855−99y^2 ∣ ⇒ ((1895)/(51))≤y^2 ≤((185)/3) ≈ 6^2 .1≤y^2 ≤7.9^2 ⇒ y=±7 ∧ x=303
$$\Leftrightarrow \\ $$$${x}=\frac{\mathrm{3}\left({y}^{\mathrm{2}} +\mathrm{355}\right)}{\mathrm{4855}−\mathrm{99}{y}^{\mathrm{2}} } \\ $$$${x}\in\mathbb{Z} \\ $$$$\Rightarrow\:\mathrm{3}\mid{y}^{\mathrm{2}} +\mathrm{355}\mid\geqslant\mid\mathrm{4855}−\mathrm{99}{y}^{\mathrm{2}} \mid \\ $$$$\Rightarrow\:\mathrm{3}\left({y}^{\mathrm{2}} +\mathrm{355}\right)\geqslant\mid\mathrm{4855}−\mathrm{99}{y}^{\mathrm{2}} \mid \\ $$$$\Rightarrow\:\frac{\mathrm{1895}}{\mathrm{51}}\leqslant{y}^{\mathrm{2}} \leqslant\frac{\mathrm{185}}{\mathrm{3}}\:\approx\:\mathrm{6}^{\mathrm{2}} .\mathrm{1}\leqslant{y}^{\mathrm{2}} \leqslant\mathrm{7}.\mathrm{9}^{\mathrm{2}} \\ $$$$\Rightarrow\:{y}=\pm\mathrm{7}\:\wedge\:{x}=\mathrm{303} \\ $$
Answered by Rasheed.Sindhi last updated on 28/Oct/24
(x/(x−3)) −(4/(y^2 −45)) = (1/(100)) let (4/(y^2 −45))=(a/(100)) ; a∈Z ⇒(x/(x−3))=(1/(100))+(a/(100)) ((y^2 −45)/4)=((100)/a) y^2 =((400)/a)+45∈Z^+ ⇒a∣400 ∧ ((400)/a)+45 is perfect square ⇒a=100⇒y=±7 (x/(x−3))=(1/(100))+(a/(100)) (x/(x−3))=(1/(100))+((100)/(100))=((101)/(100)) 100x=101x−303⇒x=303 (x,y)=(303,7),(303,−7)
$$\:\:\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{3}}\:−\frac{\mathrm{4}}{\mathrm{y}^{\mathrm{2}} −\mathrm{45}}\:=\:\frac{\mathrm{1}}{\mathrm{100}} \\ $$$${let}\:\frac{\mathrm{4}}{{y}^{\mathrm{2}} −\mathrm{45}}=\frac{{a}}{\mathrm{100}}\:;\:{a}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\frac{{x}}{{x}−\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{100}}+\frac{{a}}{\mathrm{100}} \\ $$$$\frac{{y}^{\mathrm{2}} −\mathrm{45}}{\mathrm{4}}=\frac{\mathrm{100}}{{a}} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{400}}{{a}}+\mathrm{45}\in\mathbb{Z}^{+} \\ $$$$\Rightarrow{a}\mid\mathrm{400}\:\wedge\:\frac{\mathrm{400}}{{a}}+\mathrm{45}\:{is}\:{perfect}\:{square} \\ $$$$\Rightarrow{a}=\mathrm{100}\Rightarrow{y}=\pm\mathrm{7} \\ $$$$\frac{{x}}{{x}−\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{100}}+\frac{{a}}{\mathrm{100}} \\ $$$$\frac{{x}}{{x}−\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{100}}+\frac{\mathrm{100}}{\mathrm{100}}=\frac{\mathrm{101}}{\mathrm{100}} \\ $$$$\mathrm{100}{x}=\mathrm{101}{x}−\mathrm{303}\Rightarrow{x}=\mathrm{303} \\ $$$$\left({x},{y}\right)=\left(\mathrm{303},\mathrm{7}\right),\left(\mathrm{303},−\mathrm{7}\right) \\ $$

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