If a|b+c then a|b and b|c

Suppose a|(b+c) = t for some t element of Z such that t = (k+p) for some k and p elements of Z

b+c = a(t)

b+c = a (k+p) by substitution

b+c = ak + ap by distributive property

then b = ak and c = ap

this implies a|b = k and a|c =p

thus if a|b and b|c , then a|(b+c)

please correct me if im mistaken... thnaks