I can't determine the maximum natural number which is unable to be represented as $17\times x+23\times y$

Question

The given problem is the following one.

Find the maximum natural number which is unable to be represented as $17\times x +23\times y \quad x,y\in \mathbb{N}$ .

I've been solving it till following formulas with looking the solution of the problem.

$1\leqq i \leqq 22 \quad 17 \times i \equiv r_i(mod 23)$

$17\times i\not\equiv0(mod23)$

$n(r)=22$

Therefore we can assume the problem's formula as $(1\leqq x\leqq23 \land y,q\in\mathbb{N})\quad 17\times x+23\times y =23\times q+r\prime \quad\dots(1)$ .

$\uparrow $Because if x is greater than 23,then $17\times x$ is represented as $17\times(23\times t+u)\quad (t\in \mathbb{N,u \in\{1,2,...,23\}})$

$=17\times (23\times t)+17\times u ...(2)$

and $1$st term of $(2)$ is able to be absorbed in $2$nd term of$(1)$.

And my problem begins from here.

The solution says that it is possible to be represented the all natural numbers which are greater than $17\times 23$ with $(1)$and it is impossible to be represented $17\times23$,so the answer is $17\times 23$.

Of course I can deduce that $17\times 23 $ is unable to be represented as a form of $(1)$ but I can't understand that"it is possible to be represented the all natural numbers which are greater than $17\times 23$".

How could this proven?.

Answer 1

$ 17x\!+\!23y = \color{#0a0}{n > 17(23)}\,$ is solvable for $\,x,y\ge 1,\,$ by mod $23\!:\, $ there's $\,x\equiv 17^{-1}n,\ 1\le \color{#c00}{x \le 23},\,$ so $ 17x \equiv n,\,$ so $\,17 x + 23 y = n,\,$ for $\, y\in\Bbb Z,\,$ and $\,y>0\,$ by $\,17\color{#c00}x \le 17(\color{#c00}{23})\color{#0a0}{< n}$

$ 17x\!+\!23y\: {\bf\color{#c00}=}\: 17(23)\,$ is $\rm\color{#c00}{unsolvable}$: $\, 17\mid 23y\Rightarrow 17\mid y\ $ so $\ x\!+\!23\:\!{\large \frac{y}{17}}\! = 23\,$ contra $\,x,{\large \frac{y}{17}} \ge 1$

We used: if $\,\gcd(a,b)=1\,$ then $\,a^{−1}$ exists $\!\bmod b\,$ (e.g. via Bezout gcd identity) and, furthermore $\,a\mid bc\Rightarrow a∣c\,$ (by Euclid's Lemma), for $\,a,b=17,23\,$ above. Since we used only coprimality of $\,17,23,\,$ the proof works for any coprime $\,a,b > 1,\,$ e.g. see here, which has a more geometric proof, along with citations on this well-known Frobenius Coin Problem.

Remark $\, $ A unit shift translates the above to permit $\,x,y = 0,\,$ namely $\ \ \ \ \begin{align} &\ \ \ \ \, 17\,x^{\phantom{|^|}} \ \,+\ \ \ \ 23\,y\, \ \ \ =\ \ \ \ n\qquad\quad\ \ \,{\rm for}\ \ x,y \ge 0\\[.2em] \iff\ &17(x\!+\!1) + 23(y\!+\!1) =\, n\!+\!17\!+\!23\,\ \ {\rm for}\ \ x\!+\!1,y\!+\!1\ge 1,\ \text{so by above}\\[.2em] &{\rm this\ \ is\,\ \underset{\textstyle\color{#c00}{unsolvable}}{ solvable}\:\ for}\ \ \,n\!+\!17\!+\!23\underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}> 17(23)\ \ {\rm i.e.}\ \ n \underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}> 17(23)\!-\!17\!-\!23 \end{align}$

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Alternatively we can use this inductive proof. Let $\,f(x,y)=17x+23y.\,$ The Bezout identity is $\,f(-4,3)=1.\,$ Hence if $\,f(x,y) = \color{#0a0}n\,$ then $\,f(\color{#c00}{x-4},y+3) = f(x,y)+f(-4,3) = \color{#0a0}{n+1},\,$ and if $\,\color{#c00}{x-4\le 0}\,$ then we can make it $> 0$ by further $\rm\color{#c00}{adding\ (23,-17)},\,$ since $\,f(23,-17)=0,\,$ e.g.

$391 = f(\ \ \ 0,17),\ $ so adding $1 = f(-4,3)\,$ yields
$392 = f(\color{#c00}{-4},20) = f(19,3)$ by $\rm\color{#c00}{adding\ (23,-17)}$
$393 = f(\ 15,6)$
$394 = f(\ 11,9)$
$395 = f(\ \ \ 7,12)$
$396 = f(\ \ \ 3,15)$
$397 = f(\color{#c00}{-1},18) = f(22,1)$ by $\rm\color{#c00}{adding\ (23,-17)}$
$\qquad \vdots$