For which natural numbers n do there exist nonnegative integers x and y such that... n=4x+5y? Justify your conclusion.

My conjecture/proposition:

For n=4,5,8,9,10, and 12 and beyond, there exist nonnegative integers x and y such that n=4x+5y. (0 is not a natural number)

My proof:

Let x=1, and y=0, thus 4(1)+5(0)=4 which is a natural number.

Let x=0, and y =1, thus 4(0)+5(1)=5 which is a natural number.

Let x=2, and y =0, thus 4(2)+5(0)=8 which is a natural number.

Let x=1, and y=1, thus 4(1)+5(1)=9 which is a natural number.

Let x=0, and y=2, thus 4(0)=5(2)=10 which is a natural number.

My initial step:

Assume P(12) is true. Let x=3, y=0. Thus 4(3)+5(0)=12 which is a natural number and x and y are nonnegative integers.

Inductive step:

Must prove that P(k+1)=4x+5y.

My question is, can I assume K+1=(K-3)+4 and K-3=4x+5y?

If I can then, K-3=4x+5y, then adding 4 to each side I can show that

K+1=4x+4+5y,

K+1=4(x+1)+5y since (x+1) and 5y are nonnegative integers, so are x and y.

How does this proof seem? If it isn't up to par, what am I missing?