Given-any-positive-integer-n-show-that-there-are-two-positive-rational-numbers-a-and-b-a-b-which-are-not-integers-and-which-are-such-that-a-b-a-2-b-2-a-3-b-3-a-n-b-n-are-al

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Question Number 22080 by Tinkutara last updated on 10/Oct/17

$$\mathrm{Given}\mathrm{any}\mathrm{positive}\mathrm{integer}n\mathrm{show}$$$$\mathrm{that}\mathrm{there}\mathrm{are}\mathrm{two}\mathrm{positive}\mathrm{rational}$$$$\mathrm{numbers}a\mathrm{and}b,a\ne b,\mathrm{which}\mathrm{are}\mathrm{not}$$$$\mathrm{integers}\mathrm{and}\mathrm{which}\mathrm{are}\mathrm{such}\mathrm{that}a-b,$$$$a^2-b^2,a^3-b^3,\dots ..,a^n-b^n\mathrm{are}\mathrm{all}$$$$\mathrm{integers}.$$

Commented by Rasheed.Sindhi last updated on 11/Oct/17

$$\mathrm{For}\mathrm{n}=2$$$$a-b,a^2-b^2\in \mathbb{Z}\Rightarrow a+b\in \mathbb{Z}$$$$[\because a^2-b^2=(a-b)(a+b)]$$$$\frac{7}{2}-\frac{1}{2}=3$$$${\left(\frac{7}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2=\frac{49}{4}-\frac{1}{4}=12$$$$a-b,a+b\in \mathbb{Z}\Rightarrow a^2-b^2\in \mathbb{Z};a,b\in \mathbb{Q}$$$$\mathrm{For}\mathrm{n}=3$$$$a-b\in \mathbb{Z}$$$$a^2-b^2=(a-b)(a+b)\in \mathbb{Z}$$$$\Rightarrow a+b\in \mathbb{Z}$$$$a^3-b^3=(a-b)(a^2+ab+b^2)\in \mathbb{Z}$$$$\Rightarrow a^2+ab+b^2\in \mathbb{Z}$$$$⋮$$

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