TITLE: Inductively prove that any natural number $\ge 12$ can be written as the sum of 4s and 5s
QUESTION [6 upvotes]: I can intuitively see why this is true:

Let us assume $n = \alpha \times 4 + \beta \times 5$ with $\alpha,\beta \in \mathbb{N} \cup \{0\}$.
$\forall n \in \mathbb{N} \cup \{0\}$: $n \div 4$ will yield a remainder $r \in \mathbb{N} \cup \{0\}$, $0 \le r \lt 4$.
$r$ can be 'split' into 1s, and each of those 1s can be added to one
of the 4s that go into $n$ to turn them into 5s. Therefore, if $n =
    a\times4 + r \Rightarrow n = (a-r)\times4 + r\times5$ with $a \in \mathbb{N}\cup\{0\}$.
That is the reason why this only works for $n \ge 12=3\times4$: it's readily
obvious by the pigeonhole principle, and if $a \lt 3 \Rightarrow
    \exists \ r \ | \ (a-r) \lt 0 \Rightarrow \alpha \lt 0$ which contradicts the initial assumption.

Could this give me a start for the inductive proof? I don't know if induction would be the most straightforward or elegant way to prove it, but I have to do it as an exercise (my Elementary Number Theory textbook actually suggests proving it individually for $n\in[12,16]$ and then using 16 as $n_0$). I understand how induction works but I can't think of a way to translate this problem into it. I would also like some help with formally expressing step 3 of my proof.

REPLY [4 votes]: $12=4+4+4$
$13=4+4+5$
$14=4+5+5$
$15=5+5+5$
Now, suppose $n>15$; as the inductive hypothesis you can assume that any number $m$ with $12\le m<n$ can be written as sum of fours and fives. Then $n-4>11$ can be written as sum of fours and fives, which implies the thesis also for $n$.

This is indeed a constructive approach: divide $n\ge12$ by $4$, with remainder $r$, that is, $n=4q+r$, with $0\le r<4$. Then $q\ge3$, as $n\ge12$.
If $r=0$ we are done. 
If $r=1$, $n=4(q-1)+5$
If $r=2$, $n=4(q-2)+5+5$
If $r=3$, $n=4(q-3)+5+5+5$