I would stop here, and from this point continue as

b+c= ak+ap

= a( k+ p )

= a z

where z = k+p where clearly k+p is an integer, thus a | (b + c).

I don't really follow what you did in the rest of the proof.

Hence, ak+ap is an integer b/c integers and sums and products of integers are integers.

Therefore, a|(b+c)= b+c=a*z for some integer z.

So, ak+ap= az for some integer z (by substitution)

Solve for z to prove that z is an integer:

z=k+p.

Therefore, a is divisible by b+c since b+c= a*z for some integer z.

**END OF PROOF**.