show that if p (prime integer)divise a^n then p divise also a.


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 127897 by mathocean1 last updated on 02/Jan/21

$show that if p (prime integer) divise$ $a^{n} then p divise also a .$
Commented by JDamian last updated on 03/Jan/21

$t h i s q u e s t i o n s e e m s b i z a r r e f o r m e b e c a u s e$ $if a^{n} \mod t = 0 \forall n > 1 \Leftrightarrow a \mod t = 0$
	Answered by floor(10²Eta[1]) last updated on 03/Jan/21

	$(★)$ $if n ∣ a and n ∣ b \Rightarrow n ∣ ax + by, x, y \in Z$ $n ∣ a \Rightarrow a = {nk}_{1}, k_{1} \in Z$ $n ∣ b \Rightarrow b = {nk}_{2}, k_{2} \in Z$ $ax = {nk}_{1} x and by = {nk}_{2} y$ $\Rightarrow n ∣ ax + by = n (k_{1} x + k_{2} y)$ $(◼)$ $if a \equiv b (\mod n) and c \equiv d (\mod n) so$ $ac \equiv bd (\mod n)$ $a \equiv b (\mod n) \Rightarrow n ∣ a - b$ $c \equiv d (\mod n) \Rightarrow n ∣ c - d$ $by (★) we know that$ $n ∣ c (a - b) + b (c - d) \Rightarrow n ∣ ac - bd$ $\Rightarrow ac \equiv bd (\mod n)$ $Now suppose p ∣ a^{n} but p ∤ a :$ $\Rightarrow a ≢ 0 (\mod p)$ $by (◼) a^{2} \equiv b^{2} (\mod n) (a = c, b = d)$ $by induction a^{k} \equiv b^{k} (\mod n), k \in N$ $a ≢ 0 (\mod p) \Rightarrow a^{n} ≢ 0^{n} = 0 (\mod p)$ $\Rightarrow p ∤ a^{n}, contradiction because we suppose$ $that p ∣ a^{n} . So if p ∣ a^{n} \Rightarrow p ∣ a$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum