Idea of the Proof : Existence of a & b so that (Any integer greater than 8) = 3a + 5b

Question

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.

Answer 1

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 $a, b$.

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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 $$ x_0 - t m > -1, \qquad t n + y_0 > -1\tag{eq'} $$

Answer 2

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.

Answer 3

$3a=3\cdot a+0\cdot5$ 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 $2$

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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 $1,2,4,7$

Answer 4

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.