show-that-if-p-prime-integer-divise-a-n-then-p-divise-also-a-

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