How to prove for each $n\in \mathbb{N}$ (with $n\ge12$), there exist $a_n,b_n\in \mathbb{N}\cup\{0\} $ such that $n=4a_n+5b_n$. I need to prove the following statement.

For each $n\in \mathbb{N}$ (with $n\ge12$), there exist $a_n,b_n\in \mathbb{N}\cup\{0\} $ such that $n=4a_n+5b_n$.

Please help me. Thanks.
 A: $$
12 = 4*3 + 5*0
$$
$$
13 = 4*2 + 5*1
$$
$$
14 = 4*1 + 5*2
$$
$$
15 = 4*0 + 5*3
$$
For any other $n\geq16$, we can keep adding $4$ to one of the above to get to it.  
A: hint: $n = 12 \Rightarrow a_1 = 3, b_1 = 0, n = 4a_n + 5b_n \Rightarrow n+1 = 4a_n + 5b_n + 5 - 4 = 4(a_n-1) + 5(b_n+1) = 4a'_n+5b'_n$. It can be done by induction, and you can reason a little bit on the $a_n-1 \geq 0$.
A: Hint:
Can you see why it would suffice to show that you can get $12, 13, 14$, and $15$ in such a way?

If you are interested, the generalized version of this problem is known as the Frobenius coin problem (albeit stated slightly differently).  In general, if you have two coprime numbers $x$ and $y$, then you are able to generate every natural number $N > xy - x - y$ via positive linear combinations of $x$ and $y$. See the link here:
http://en.wikipedia.org/wiki/Coin_problem
At the risk of venturing too off-topic, a famous, particularly amusing case of the problem for $3$ numbers is known as the "Chicken McNugget Problem".
A: If $n$ is divisible by $4$, you can set $a_n = n/4$ and $b_n = 0$. If $n$ leaves remainder $1$ upon division by $4$, you may set $a_n = (n - 5)/4$ and $b_n = 1$. If the remainder is $2$, set $a_n = (n - 10)/4$ and $b_n = 2$. Finally if the remainder is $3$, set $a_n = (n - 15)/4$ and $b_n = 3$. Note that in this final case $a_n \ge 0$ since $n \ge 15$: neither $12$, $13$, nor $14$ leave remainder $3$ upon division by $4$.
A: Any n can be written as
5 mod n =0, 1, 2,3 or 4
For 0: a=0.
for 1: The number can be written as some x*5 +1. Note that the smallest n in this case is greater than 15 always.
It can be written as (n-16) +(15+1)=(n-15)+16.
 Here (n-16) is divisible by 5 because (n-1) is divisible by 5 and 15 is divisible by 5, hence thier sum is divisible by 5.
For 2: Similarly, this can be written as (n-12)+12, where (n-12) will be divisible by 5.
For 3: Write (n-8)+8. (n-8) is divisible by 5.
For 4: Write (n-4)+4. (n-4) is divisible by 5.
A: Induction Method:
Let $P(n)$ be the statement there exists there exist $a_n,b_n\in \mathbb{N}\cup\{0\} $ such that $n+11=4a_n+5b_n$.
It is easy to see that $P(1)$ is true.
Now let $n\in \mathbb{N}$ be such that $P(n)$ is true. So, there exist $a_n,b_n\in \mathbb{N}\cup\{0\} $ such that $n+11=4a_n+5b_n$.
Observe that $(n+1)+11=4(a_n-1)+5(b_n+1)$.
Case 1: Suppose $a_n\ge 1$. 
Then $a_n-1,b_n+1\in \mathbb{N}\cup\{0\}$. Therefore $P(n+1)$ is true in this case.
Case 2: Suppose $a_n\ge 0$. 
Then $n+11=5b_n$ and $b_n\ge 3$. Therefore $(n+1)+11=5b_n+1=5(b_n-3)+4.4$ . Since $b_n-3,4\in \mathbb{N}\cup\{0\}$ we have that $P(n+1)$ is true.
Then by the Principle of Mathematical Induction, for each $n\in \mathbb{N}$, there exist $a_n,b_n\in \mathbb{N}\cup\{0\} $ such that $n+11=4a_n+5b_n$. 
Thus for each $n\in \mathbb{N}$ (with $n\ge12$), there exist $a_n,b_n\in \mathbb{N}\cup\{0\} $ such that $n=4a_n+5b_n$. 
