HW1 Ch0 24 MA453Fall2008walther - Rhea

Let a₁=a₁ and a₂a₃...a_n=b₁

If p is a prime and divides a₁a₂a₃...a_n, then p divides a₁b₁

If p is a prime that divides a₁b₁, then p divides a₁ or b₁

Let's say p does not divide a₁, then gcd(p,a₁)=1

This means that there exists x and y for which the equation xp+ya₁=1 holds

Let's multiply both sides of this equation by b₁:

xpb₁+ya₁b₁=b₁

By induction, p divides a₁b₁ and let a₁b₁=kp. Let's divide the equation above by p:

xb₁+yk=b₁/p

If the LHS of the equation can be divided by p, the RHS of the equation can be divided by p also. Then, b₁ can be divided by p.

Next, we let b₂=a₃a₄...a_n and repeat the process above. Eventually, we will find the a_i for some i, which can be divided by p.

-Ozgur