If p|ab and (a,p)= 1 ax+py = 1 ⇒ axb+pyb=b ∴ p|b and p|a by symmetry.. (case if (b,p)= 1)
Asked
Active
Viewed 77 times
1 Answers
1
It might be "and" but it is not always so.
Consider this example:
$2$ is prime, and $2$ divides $(4\times 15)$.
Now $2$ does divide $4$, but $2$ does not divide $15$.
The point is: a prime number does not have any factors beside itself and 1. Thus if a prime number is a factor of a product of two natural numbers, then it must be a factor of at least one among them.
Graham Kemp
- 133,231
Proof: I don't know if this answers my question, but i knew that let a natural and p a prime, (a,p) = 1 or p|a
Proof: suppose by absurd that gcd(a,p) = d > 1 we have d|a and d|p, but p is prime, therefore d = p or d = 1, we assume d >1, therefore d = p
∴ if d = p ⇒ p|a
And we can aply this Lemma and both numbers, can't?
– Kayan Tchian Jan 24 '20 at 01:34