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Question Number 159925 by Rasheed.Sindhi last updated on 22/Nov/21

$$\mathrm{Find}\mathrm{out}\mathrm{pairs}\mathrm{of}\mathrm{numbers}(\mathrm{a},\mathrm{b})(\mathrm{as}$$$$\mathrm{many}\mathrm{as}\mathrm{you}\mathrm{can})\mathrm{such}\mathrm{that}:$$$$\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}},\mathrm{a}+\mathrm{b},{\mathrm{a}}^2+{\mathrm{b}}^2\in \mathbb{P}$$

Answered by nikif99 last updated on 23/Nov/21

$$Pairs(a,b)upto121:$$$$(4,9),(4,25),(9,64),(25,64),$$$$(36,121),(49,100),(81,100).$$

Commented by Rasheed.Sindhi last updated on 23/Nov/21

$$\mathbb{T}\mathrm{han}\mathbb{k}\mathrm{s}\mathrm{sir}!$$

Commented by Rasheed.Sindhi last updated on 23/Nov/21

$${\mathrm{What}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{n}\mathrm{for}$$$$\mathrm{which}\mathrm{a}+\mathrm{b},{\mathrm{a}}^2+{\mathrm{b}}^2,{\mathrm{a}}^3+{\mathrm{b}}^3,...,{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}\in \mathbb{P}$$

Commented by Rasheed.Sindhi last updated on 23/Nov/21

$$\mathrm{Or}\mathrm{if}\mathrm{m}<\mathrm{n},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{value}$$$$\mathrm{of}\mathrm{n}-\mathrm{m}\mathrm{for}\mathrm{which}$$$${\mathrm{a}}^{\mathrm{m}}+{\mathrm{b}}^{\mathrm{m}},{\mathrm{a}}^{\mathrm{m}+1}+{\mathrm{b}}^{\mathrm{m}+1},{\mathrm{a}}^{\mathrm{m}+2}+{\mathrm{b}}^{\mathrm{m}+2},...,{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}\in \mathbb{P}$$

Commented by nikif99 last updated on 23/Nov/21

$$Ifyouomitthedemand\sqrt{a}+\sqrt{b}\in \mathbb{P},$$$$thenumberofacceptablepairs$$$$increaseenormously.$$$$Nevertheless,max\left(n\right)seemstobe2.$$$$P.S.Anotherpairtill200:(100,169).$$

Commented by Rasheed.Sindhi last updated on 23/Nov/21

$$\mathcal{T}hanksalotsir!{You}^{\prime }rerightbecause$$$$a^3+b^{3}isalwayscomposite.$$$$\mathcal{T}heclaimshouldbe:$$$$a+b,a^2+b^2,a^4+b^4,a^8+b^8,...\in \mathbb{P}$$$$Yourprovidedexamplesare$$$$successfulupto3levels.Forexample$$$$Ifinsteadof(100,169)ifwe$$$$consider(a,b)=(\sqrt{100},\sqrt{169})=(10,13)$$$$10+13=23\in \mathbb{P}$$$${10}^2+{13}^2=269\in \mathbb{P}$$$${10}^4+{13}^4=38561\in \mathbb{P}$$$$\mathbb{T}\mathrm{han}\mathbb{k}\mathrm{s}\mathrm{for}\mathrm{attention}!$$

Commented by nikif99 last updated on 23/Nov/21

$$Excellentpost!Thankyoutoo.$$

Commented by Rasheed.Sindhi last updated on 24/Nov/21

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