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Question Number 122329 by mathocean1 last updated on 15/Nov/20

$$findn\in \mathbb{N}suchthat$$$$5^{2n}+5^n\equiv 0\left[13\right]$$

Answered by 676597498 last updated on 15/Nov/20

$$letx=5^n$$$$\Rightarrow x^2+x=0mod\left(13\right)$$$$\Rightarrow {(x+\frac{1}{2})}^2-\frac{1}{4}=0mod\left(13\right)$$$$\Rightarrow {(x+\frac{1}{2})}^2=\frac{1}{4}mod\left(13\right)$$$$\Rightarrow x+\frac{1}{2}=\frac{1}{2}mod\left(13\right)orx+\frac{1}{2}=-\frac{1}{2}mod\left(13\right)$$$$\Rightarrow x=0mod\left(13\right)orx=-1mod\left(13\right)$$$$butx=5^n$$$$5^n=0mod\left(13\right)$$$$\Rightarrow n={log}_5\left(0mod\left(13\right)\right)$$$$or$$$$5^n=-1mod\left(13\right)=12$$$$\Rightarrow n={log}_{5}12$$

Commented by mindispower last updated on 16/Nov/20

$$x\in \mathbb{N}$$

Answered by MJS_new last updated on 15/Nov/20

$$5^{2n}+5^n=13m;m\in \mathbb{N}$$$$5^n(5^n+1)=13m$$$$5^n\ne 13m\left[\mathrm{obviously}\right]$$$$\Rightarrow 5^n=13m-1$$$$\mathrm{trying}\mathrm{the}\mathrm{first}\mathrm{few}n\mathrm{we}\mathrm{get}$$$$n=4k+2$$

Answered by floor(10²Eta[1]) last updated on 16/Nov/20

$$5^{\mathrm{n}}(5^{\mathrm{n}}+1)\equiv 0\left(\mathrm{mod}13\right)$$$$5^{\mathrm{n}}\equiv 0\left(\mathrm{mod}13\right)\vee 5^{\mathrm{n}}\equiv 12\left(\mathrm{mod}13\right)$$$$\Rightarrow 5^{\mathrm{n}}\equiv 12\left(\mathrm{mod}13\right)$$$$\mathrm{n}=2\Rightarrow 5^{\mathrm{n}}=5^2\equiv 12\left(\mathrm{mod}13\right)$$$$5^{\mathrm{n}-2}\equiv 1\left(\mathrm{mod}13\right)$$$$5^{\mathrm{a}}\equiv 1\left(\mathrm{mod}13\right)\Rightarrow \mathrm{a}\mid \phi \left(13\right)=12\Rightarrow \mathrm{a}=4\therefore \mathrm{a}=4\mathrm{k}$$$$\Rightarrow \mathrm{n}-2=4\mathrm{k}\Rightarrow \mathrm{n}=4\mathrm{k}+2,\mathrm{\forall }\mathrm{k}⩾0$$

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