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Category: Number Theory

Find-all-positive-integers-n-such-that-n-1-divides-n-2-1-

Question Number 217030 by ArshadS last updated on 27/Feb/25 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\:\mathrm{n}\:\:\mathrm{such}\:\mathrm{that}\:\: \\ $$$$\:\mathrm{n}\:+\:\mathrm{1}\:\:\mathrm{divides}\:\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$ Answered by issac last updated on 27/Feb/25 $$\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\frac{{n}+\mathrm{1}}{\left(\mathrm{1}+{n}\boldsymbol{{i}}\right)\left(\mathrm{1}−{n}\boldsymbol{{i}}\right)} \\…

Find-all-positive-integers-n-such-that-n-divides-2-n-1-

Question Number 217040 by ArshadS last updated on 27/Feb/25 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\:\mathrm{n}\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\mathrm{n}\:\:\mathrm{divides}\:\:\mathrm{2}^{{n}} \:+\:\mathrm{1}.\:\: \\ $$ Answered by Ghisom last updated on 27/Feb/25 $$\mathrm{one}\:\mathrm{group}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{is}\:{n}=\mathrm{3}^{{k}} \wedge{k}\in\mathbb{N} \\…

Find-all-three-digit-numbers-n-such-that-1-n-is-divisible-by-the-sum-of-its-digits-2-n-is-a-perfect-square-

Question Number 216912 by ArshadS last updated on 26/Feb/25 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{three}-\mathrm{digit}\:\mathrm{numbers}\:{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{1}.\:{n}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{the}\:\mathrm{sum}\:\:\mathrm{of}\:\:\mathrm{its}\:\:\mathrm{digits}. \\ $$$$\mathrm{2}.\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Let-p-be-a-prime-number-greater-than-3-Prove-that-p-2-1-is-always-divisible-by-24-

Question Number 216875 by ArshadS last updated on 23/Feb/25 $$\mathrm{Let}\:\:\mathrm{p}\:\:\mathrm{be}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that}\:\:\mathrm{p}^{\mathrm{2}} −\:\mathrm{1}\:\: \\ $$$$\mathrm{is}\:\:\mathrm{always}\:\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{24}. \\ $$ Answered by maths2 last updated on 23/Feb/25 $$\left({p}−\mathrm{1}\right)\left({p}+\mathrm{1}\right) \\ $$$$\mathrm{3}\mid\left({p}−\mathrm{1}\right)\left({p}+\mathrm{1}\right);{since}\:{p}\equiv\mathrm{1},\mathrm{2}\left[\mathrm{3}\right]…