find-n-N-such-that-5-2n-5-n-0-13-

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

f i n d n \in N s u c h t h a t

5^{2 n} + 5^{n} \equiv 0 [13]

Answered by 676597498 last updated on 15/Nov/20

l e t x = 5^{n}

\Rightarrow x^{2} + x = 0 m o d (13)

\Rightarrow {(x + \frac{1}{2})}^{2} - \frac{1}{4} = 0 m o d (13)

\Rightarrow {(x + \frac{1}{2})}^{2} = \frac{1}{4} m o d (13)

\Rightarrow x + \frac{1}{2} = \frac{1}{2} m o d (13) o r x + \frac{1}{2} = - \frac{1}{2} m o d (13)

\Rightarrow x = 0 m o d (13) o r x = - 1 m o d (13)

b u t x = 5^{n}

5^{n} = 0 m o d (13)

\Rightarrow n = {l o g}_{5} (0 m o d (13))

o r

5^{n} = - 1 m o d (13) = 12

\Rightarrow n = {l o g}_{5} 12

Commented by mindispower last updated on 16/Nov/20

x \in N

Answered by MJS_new last updated on 15/Nov/20

5^{2 n} + 5^{n} = 13 m; m \in N

5^{n} (5^{n} + 1) = 13 m

5^{n} \neq 13 m [obviously]

\Rightarrow 5^{n} = 13 m - 1

trying the first few n we get

n = 4 k + 2

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

5^{n} (5^{n} + 1) \equiv 0 (\mod 13)

5^{n} \equiv 0 (\mod 13) \lor 5^{n} \equiv 12 (\mod 13)

\Rightarrow 5^{n} \equiv 12 (\mod 13)

n = 2 \Rightarrow 5^{n} = 5^{2} \equiv 12 (\mod 13)

5^{n - 2} \equiv 1 (\mod 13)

5^{a} \equiv 1 (\mod 13) \Rightarrow a ∣ φ (13) = 12 \Rightarrow a = 4 ∴ a = 4 k

\Rightarrow n - 2 = 4 k \Rightarrow n = 4 k + 2, \forall k ⩾ 0