Question-211373

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Question Number 211373 by RojaTaniya last updated on 07/Sep/24

Answered by BaliramKumar last updated on 07/Sep/24

$$\mathrm{82} \\ $$

Commented by RojaTaniya last updated on 07/Sep/24

$${kindly}\:{provide}\:{solution}. \\ $$

Answered by Rasheed.Sindhi last updated on 07/Sep/24

n^2 +1≡0(mod 269) n^2 ≡−1(mod 269 n^2 ≡−1+269(25)(mod 269) n^2 ≡6724(mod 269) n≡82(mod 269)

$${n}^{\mathrm{2}} +\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{269}\right) \\ $$$${n}^{\mathrm{2}} \equiv−\mathrm{1}\left({mod}\:\mathrm{269}\right. \\ $$$${n}^{\mathrm{2}} \equiv−\mathrm{1}+\mathrm{269}\left(\mathrm{25}\right)\left({mod}\:\mathrm{269}\right) \\ $$$${n}^{\mathrm{2}} \equiv\mathrm{6724}\left({mod}\:\mathrm{269}\right) \\ $$$${n}\equiv\mathrm{82}\left({mod}\:\mathrm{269}\right) \\ $$

Commented by RojaTaniya last updated on 07/Sep/24

$${Sir},\:{how}\:{decide}\:\mathrm{269}\left(\mathrm{25}\right)? \\ $$

Commented by Rasheed.Sindhi last updated on 07/Sep/24

Searching for such value of m for which −1+269m is perfect square. So m=25

$${Searching}\:{for}\:{such}\:{value}\:{of}\:{m}\:{for} \\ $$$${which}\:−\mathrm{1}+\mathrm{269}{m}\:{is}\:{perfect}\:{square}. \\ $$$${So}\:{m}=\mathrm{25} \\ $$

Commented by BaliramKumar last updated on 08/Sep/24

n = 269k + 82 {k = 0, 1, 2, 3, .........}

$${n}\:=\:\mathrm{269}{k}\:+\:\mathrm{82}\:\:\:\:\:\:\:\:\:\:\:\:\left\{{k}\:=\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:………\right\} \\ $$

Answered by Frix last updated on 08/Sep/24

n, k ∈Z: n=269k±82 n, k ∈N: n=269k+82 n=269k+187 ⇒ min (n∈N) =82 [N={0, 1, 2, ...}]

$${n},\:{k}\:\in\mathbb{Z}:\:{n}=\mathrm{269}{k}\pm\mathrm{82} \\ $$$${n},\:{k}\:\in\mathbb{N}: \\ $$$${n}=\mathrm{269}{k}+\mathrm{82} \\ $$$${n}=\mathrm{269}{k}+\mathrm{187} \\ $$$$\Rightarrow\:\mathrm{min}\:\left({n}\in\mathbb{N}\right)\:=\mathrm{82} \\ $$$$ \\ $$$$\left[\mathbb{N}=\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:…\right\}\right] \\ $$

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