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Question Number 211753 by alcohol last updated on 19/Sep/24

find all (n,m) such that ((n^2 −m)/(m^2 −n)) ∈ Z

$${find}\:{all}\:\left({n},{m}\right)\:{such}\:{that}\:\frac{{n}^{\mathrm{2}} −{m}}{{m}^{\mathrm{2}} −{n}}\:\in\:\mathbb{Z} \\ $$

Answered by Frix last updated on 20/Sep/24

There are infinitely many solutions. 3 ways to find some: 1. Let m∈Z\{0, 1}∧n=m ⇒ ((n^2 −m)/(m^2 −n))=1 2. Let m∈Z∧n=−(m+1) ⇒ ((n^2 −m)/(m^2 −n))=1 3. Let m∈Z∧n=−m(m^3 −m−1) ⇒ ((n^2 −m)/(m^2 −n))=m^4 −2m^2 −m+1

$$\mathrm{There}\:\mathrm{are}\:\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solutions}. \\ $$$$\mathrm{3}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{find}\:\mathrm{some}: \\ $$$$\mathrm{1}.\:\mathrm{Let}\:{m}\in\mathbb{Z}\backslash\left\{\mathrm{0},\:\mathrm{1}\right\}\wedge{n}={m}\:\Rightarrow\:\frac{{n}^{\mathrm{2}} −{m}}{{m}^{\mathrm{2}} −{n}}=\mathrm{1} \\ $$$$\mathrm{2}.\:\mathrm{Let}\:{m}\in\mathbb{Z}\wedge{n}=−\left({m}+\mathrm{1}\right)\:\Rightarrow\:\frac{{n}^{\mathrm{2}} −{m}}{{m}^{\mathrm{2}} −{n}}=\mathrm{1} \\ $$$$\mathrm{3}.\:\mathrm{Let}\:{m}\in\mathbb{Z}\wedge{n}=−{m}\left({m}^{\mathrm{3}} −{m}−\mathrm{1}\right)\:\Rightarrow \\ $$$$\:\:\:\:\:\frac{{n}^{\mathrm{2}} −{m}}{{m}^{\mathrm{2}} −{n}}={m}^{\mathrm{4}} −\mathrm{2}{m}^{\mathrm{2}} −{m}+\mathrm{1} \\ $$

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