<s>[INST] 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.
 [/INST] $3a=3\cdot a+0\cdot5$  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 $1,2,4,7$ 
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