Idea of the Proof : Existence of a & b so that (Any integer greater than 8) = 3a + 5b 
Claim: Prove that for every integer $n \geq 8$, there exist nonnegative integers $a$ and $b$ such that $n = 3a + 5b.$
Proclaimed solution : Let $n ∈ \mathbb{Z}$ with $n ≥ 8.$ $\text{ Then } n = 3q, \text{ where } q ≥ 3, \quad \text{ or } n = 3q+1, \text{ where } q ≥ 3, \quad \text{ or } n = 3q+2, \text{ where } q ≥ 2.$
  We consider these three cases. (Rest of proof omitted)

I do not apprehend the above parts of the proof BEFORE (Rest of proof omitted). However, I understand and completed the rest of the proof that I omitted. I recognise that it uses the result that division of any integer by $n$ must yield a remainder of $0, 1, 2, ..., n - 2, 
\text{ or } n - 1$. 
$1.$ However, what is the idea or strategy of the proof? 
$2.$ What motivates or propounds the claim?
$3.$ Is there a more natural or easier proof?
I referenced 1. Source: P102, Problem 4.9 on P101 (related to P90, Result 4.8) of Mathematical Proofs, 2nd ed by Chartrand et al.
 A: More generally, given coprime positive integers $n, m$, every integer $z$ greater than or equal to $(n-1)(m-1)$ can be written in the form
$$
z = a n + b m
$$
for non-negative integers $a, b$. 

Here's a proof
First find integers $x_0, y_0$ such 
$z = x_0 n - y_0 m.$
Now we know that the solutions $x, y$ of
$$
z = x n + y m
$$
are of the form
$$
x = x_0 - t m, \qquad y = t n + y_0,
$$
for some $t$.
For which $c$ there is a $t$ such that 
$$x_0 - t m \ge 0, \qquad t n + y_0 \ge 0?\tag{eq}$$ 
We want
$$
\frac{x_0}{m} \ge t \ge \frac{y_0}{n}.
$$
There is definitely such a $t$ if 
$$
1 \le \frac{x_0}{m} - \frac{y_0}{n} = \frac{x_0 n - y_0 m}{mn} = \frac{z}{mn},
$$
or
$z \ge m n$.
To get the best-possible estimate $z \ge (n-1)(m-1)$, replace (eq) with 
$$
x_0 - t m > -1, \qquad t n + y_0 > -1\tag{eq'}
$$
A: I can't talk of an incomplete proof, but I'd go as follows (hints):
$$\begin{align*}(1)&\;\;\min\{3a+5b\;;\;3a+5b> 0\;,\;(a,b)\in\Bbb N^2\}=8\\
(2)&\;\;\text{Induction on }\;\;n:\;\;n+1=3a+5b+1=\\
&\;\;=3a+5b+3\cdot 2+5\cdot(-1)=3\cdot(a+2)+5\cdot(b-1)\end{align*}$$
Now all is left is to show  $\,b-1\ge 0\;\ldots$and you may want to check some different cases.
A: $3a=3\cdot a+0\cdot5$  which needs $a\ge0\implies 3a\ge 0$
$3a+1=3(a-3)+2\cdot5$  which needs $a\ge3\implies 3a+1\ge 10$ so disallows $1,4,7$
$3a+2=3(a-1)+5\cdot5$  which needs $a\ge1\implies 3a+2\ge 5$  so disallows $2$

Iterating with $5$
$5b=5\cdot b+0\cdot3$ which needs $b\ge0\implies 5b\ge 0$
$5b+1=5(b-1)+2\cdot3$ which needs $b\ge1\implies 5b+1\ge 6$
$5b+3=5\cdot b+1\cdot3$ which needs $b\ge0\implies 5b+3\ge 3$
$5b+2=5(b-2)+3\cdot4$ which needs $b\ge2\implies 5b+2\ge 12$
$5b+4=5(b-1)+3\cdot3$  which needs $b\ge1\implies 5b+4\ge 9$
So, the only numbers those are not representable are $1,2,4,7$ 
A: Let $n=3a + 5b$. This leads to:
$$ n\equiv 3a + 5b \equiv 5b \pmod 3$$
Now we have three possibilities:
Case 1:
$$ n \equiv 0 \pmod 3 \implies 5b \equiv 0 \pmod 3$$
$$ 5b \equiv 15 \pmod 3$$
$$ b \equiv 3 \equiv 0 \pmod 3$$
This means that if $ n \equiv 0 \pmod 3$, we can fix $b=0$ and $a=\frac{n}{3}$, which is an integer, because of the congruence relation.
Case 2:
$$ n \equiv 1 \pmod 3 \implies 5b \equiv 1 \pmod 3$$
$$ 5b \equiv 10 \pmod 3$$
$$ b \equiv 2 \pmod 3$$
This means that if $ n \equiv 1 \pmod 3$, we can fix $b=2$ and $a=\frac{n-10}{3}$, which is an integer, because $n \equiv 10 \pmod 3$, also $n-10$ can't be negative number, because $n \geq 8$ and the next number of the form $3k+1$ is 10.
Case 3:
$$ n \equiv 2 \pmod 3 \implies 5b \equiv 2 \pmod 3$$
$$ 5b \equiv 5 \pmod 3$$
$$ b \equiv 1 \pmod 3$$
This means that if $ n \equiv 1 \pmod 3$, we can fix $b=1$ and $a=\frac{n-5}{3}$, which is an integer, because $n \equiv 5 \pmod 3$.
Because we exhausted all the cases, we proved that there is an non-negative integer solutions $(a,b)$, for every $n \geq 8$
Q.E.D.
