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3k-4-n-2-k-n-N-Find-all-n-numbers-




Question Number 224691 by lockedin last updated on 27/Sep/25
3k+4=n^2 . k,n ∈N  Find all n numbers .
$$\mathrm{3}{k}+\mathrm{4}={n}^{\mathrm{2}} .\:{k},{n}\:\in\mathbb{N} \\ $$$${Find}\:{all}\:{n}\:{numbers}\:. \\ $$
Answered by Rasheed.Sindhi last updated on 27/Sep/25
3k+4=n^2 . k,n ∈N  n^2 −4=3k  (n−2)(n+2)=3k  n−2=3,6,9,... ∣ n+2=3,6,9,...  n=5,8,11,...   ∣ n=1^(×) ,4,7,...  n=5+3m ∣ n=4+3m ; m=0,1,2,3,...  Or  n=2+3m or n=1+3m for n∈N
$$\mathrm{3}{k}+\mathrm{4}={n}^{\mathrm{2}} .\:{k},{n}\:\in\mathbb{N} \\ $$$${n}^{\mathrm{2}} −\mathrm{4}=\mathrm{3}{k} \\ $$$$\left({n}−\mathrm{2}\right)\left({n}+\mathrm{2}\right)=\mathrm{3}{k} \\ $$$${n}−\mathrm{2}=\mathrm{3},\mathrm{6},\mathrm{9},…\:\mid\:{n}+\mathrm{2}=\mathrm{3},\mathrm{6},\mathrm{9},… \\ $$$${n}=\mathrm{5},\mathrm{8},\mathrm{11},…\:\:\:\mid\:{n}=\overset{×} {\mathrm{1}},\mathrm{4},\mathrm{7},… \\ $$$${n}=\mathrm{5}+\mathrm{3}{m}\:\mid\:{n}=\mathrm{4}+\mathrm{3}{m}\:;\:{m}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},… \\ $$$${Or} \\ $$$${n}=\mathrm{2}+\mathrm{3}{m}\:{or}\:{n}=\mathrm{1}+\mathrm{3}{m}\:{for}\:{n}\in\mathbb{N} \\ $$

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