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Question Number 212926 by efronzo1 last updated on 27/Oct/24

$$\mathrm{find}\mathrm{integers}\mathrm{x},\mathrm{y}\mathrm{such}\mathrm{that}$$$$\frac{\mathrm{x}}{\mathrm{x}-3}-\frac{4}{{\mathrm{y}}^2-45}=\frac{1}{100}$$

Answered by Frix last updated on 27/Oct/24

$$\iff$$$$x=\frac{3(y^2+355)}{4855-99y^2}$$$$x\in \mathbb{Z}$$$$\Rightarrow 3\mid y^2+355\mid ⩾\mid 4855-99y^2\mid$$$$\Rightarrow 3(y^2+355)⩾\mid 4855-99y^2\mid$$$$\Rightarrow \frac{1895}{51}⩽y^2⩽\frac{185}{3}\approx 6^2.1⩽y^2⩽7.9^2$$$$\Rightarrow y=\pm 7\wedge x=303$$

Answered by Rasheed.Sindhi last updated on 28/Oct/24

$$\frac{\mathrm{x}}{\mathrm{x}-3}-\frac{4}{{\mathrm{y}}^2-45}=\frac{1}{100}$$$$let\frac{4}{y^2-45}=\frac{a}{100};a\in \mathbb{Z}$$$$\Rightarrow \frac{x}{x-3}=\frac{1}{100}+\frac{a}{100}$$$$\frac{y^2-45}{4}=\frac{100}{a}$$$$y^2=\frac{400}{a}+45\in {\mathbb{Z}}^{+}$$$$\Rightarrow a\mid 400\wedge \frac{400}{a}+45isperfectsquare$$$$\Rightarrow a=100\Rightarrow y=\pm 7$$$$\frac{x}{x-3}=\frac{1}{100}+\frac{a}{100}$$$$\frac{x}{x-3}=\frac{1}{100}+\frac{100}{100}=\frac{101}{100}$$$$100x=101x-303\Rightarrow x=303$$$$(x,y)=(303,7),(303,-7)$$

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