Never mind it did turn out to be easy.

it is sufficient to show that a | b+c and a | b --> a | c

suppose a | b+c and a | b

b+c=ak for some integer k

b=ar for some integer r

ar+c=ak

c=ak-ar

c=a(k-r)

(k-r) is an integer by closure properties

...

by contradiction

b=ak for some integer k

b+c=ar for some integer r

...

c=a(r-k)

...

a divides c contradicts ~(a divides c)