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