A simplre Problem with prime numbers

Question: Suppose p is a prime number......p is a factor of ab.....p doesn't                        divide a then prove that p is a factor of b.

Solution 1:                     Manual Solution

Since P is factor of ab....so we can write..

ab = k.P   here k is positive integer---------(1)

 

Now P is not a factor of a

Lets us assume x belong to R such that a+x is divided by P

 

So we can write  m.P = (a+x)  where  m is positive integer ------------(2)

so a = m.P - x 

 

Put in first

 

(m.P-x)(b) = k.P

 

 

m.P.b - bx  = k.P

 

from here GCD(P.b , b) = P

 

Hence P is also a factor of b

Solution 2:                                      Bezout's theorem

 Suppose p does not divide a, then gcd(p,a) =1.

 

Then by Bezout's theorem we can find integers m and n such that pm+an =1.

 

Multiply by b to obtain pmb  + abn = b

 

Both summands on LHS are divisible by p and hence LHS is divisible by p. Which means RHS=b is divisible by p.