Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $\mathcal{S}=\{ax+by\mid x,y\ge0 \text{ and } x,y\in \mathbb{Z}\}$.

Question

Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $$\mathcal{S}=\{ax+by \mid x,y\geq0 \hspace{0.25cm} \text{ and } \hspace{0.25cm} x,y\in \mathbb{Z}\}.$$

Since $a,b\in \mathbb{N}$ and $\gcd(a,b)=1$, then there exists $x',y'\in\mathbb{Z}$ such that $$ax'+by'=1.$$ Let $\alpha\in\mathcal{S}$. Then $$\alpha=ax+by$$ where $a,b\in \mathbb{N}$ with $\gcd(a,b)=1$ and $x,y\in \mathbb{Z}$ with $x,y\geq 0$.

If we were to multiply the first equation by $\alpha$ we will get $$\alpha ax'+\alpha by'=\alpha.$$ But $\alpha=ax+by$. So, $$\alpha ax'+\alpha by'=\alpha=ax+by.$$ We can rewrite this equation as follows: $$a(\alpha x'-x)=b(y-\alpha y').$$ Than $a(\alpha x'-x)| b$ or $a(\alpha x'-x)| (y-\alpha y')$ or $a(\alpha x'-x)| b(y-\alpha y')$.

From here I am unsure how to proceed. Any tips

Answer 1

I'm not certain about your divisibility conditions. It seems like only the third will hold. I proceed slightly differently.

Let's assume, without loss of generality, that $1<a<b$. Then we can construct an array of non-negative integers the is $a$ units wide and $b$ units long. This gives us a way to count all numbers from $1$ to $ab$. It is clear that all positive multiples of $a$ will be in $S$.

Ask yourself this: what happens to multiples of $b$? For $1\leq k\leq a-1$ we have $bk$ will land in some nonzero congruence $\mod a$. In fact, each will be distinct (by coprimality).

In addition, every element beneath $bk$ will be in $S$, since you just add multiples of $a$ to get to the next row. Once you get to $(a-1)b$, you have all congruence class filled $\mod a$. If you subtract $a$ from this value ($ab-b-a$), you have the last element that is not in $S$.

Every natural number greater than or equal to $ab-b-a+1$ will be in $S$! Let's see this with $a=3$ and $b=5$

\begin{align*} 1\hspace{1cm}2\hspace{1cm}s_1\\ 4\hspace{.95cm}s_2\hspace{.9cm}s_3\\ 7\hspace{.95cm}s_4\hspace{.9cm}s_5\\ s_6\hspace{.85cm}s_7\hspace{.9cm}s_8\\ s_9\hspace{.8cm}s_{10}\hspace{.7cm}s_{11} \end{align*} where the $s$ values are 3, 5, 6, 8, 9, 10, 11, 12, 13, 14, and 15. Once we get to $3\cdot5-3-5+1=8$, we intersect all integers.

Answer 2

$S \subset \mathbb N$ (let's include $0$ in our natural numbers) clearly. $a\mathbb N \subset S$ and $b\mathbb N \subset S$ so if $a = 1$ or $b =1$, $S = \mathbb N$.

So it suffices to assume $a > 1$ and $b > 1$.

Okay some bold claims:

1) $n = a(b-1) + b(-1) = a(-1) + b(a-1)=ab-a-b \not \in S$.

2) For any $n > ab-a-b$, $n \in S$.

And as $S \subset \mathbb N$, follows that $S = \{ak + bj| 0\le k <b, 0 \le j < a\} \cup \{n > ab - a - b| n \in \mathbb N\}$.

...

Clain 3: If $n = ax + by$ then $n = aw + bv \iff a = x + kb; b = x-ka$ for some $k \in \mathbb Z$.

Pf: $a(x+kb) + b(x - ka) = ax + by +akb - akb = ax + by$

Let $aw + bv = ax + by$ and let $m = w-x$. Then $ax + by = aw + bv = a(x+m) + bv =ax + bv+am = ax + b(v+\frac{am}b)$

So $y = v + \frac{am}b$ is an integer. So $b|am$ but $\gcd(a,b) = 1$ so $b|m$. So $m = kb$ and $w = x + kb$ and $v = y - \frac{akb}b =y-ak$.

So now the first claim:

Let $n = a(b-1) +b(-1) = a(w) + bv$. Then $w = b-1 + kb$ and $v= -1 - ka$. If $k \ge 0$ then $-1-ka \le -1 < 0$. If $k > 0$ then $k \ge 1$ and $w = b-1 + kb \le -1 < 0$.. So there are no $x,y \ge 0$ so that $ax +by = n$

Now the second claim.

Well, first claim 4: There exist $0 < W < b$ and $0 < V < a$ so that $aW - bV = 1$

By euclids algorithm we know there exist $x, y$ so that $ax + by = 1$. As $a,b > 0$ it's it must be that one of $x,y$ is positive and the other is negative. Now by archimedian principle we can find an integer $k$ so that $0 \le x + kb < b$. Let $W = x + kb$ and let $V = ka - y$ so $aW - bV = a(x + kb) + b(y - ka) = 1$

If $W = 0$ then $1=aW - bV = -bV$ which is impossible as $b \not \mid 1$.

So $0 < W= x+kb < b$ so

$\frac{-x}b < k < \frac {b-x}b = 1 - \frac xb$

$- \frac{xa}b - y < ka - y = V < a - \frac {xa}b$

So $V < a$ and if $V \le 0$ then $aW - bV > aW > 1$. So $0 < V$.

Okay, now the second claim.

Let $n = ab -a -b + 1= ab -a -b +aW -bV$

$=a(W -1) + b(a-1-V)$

$W \ge 1$ so $W-1\ge 0$. $a > V$ so $a-V > 0$ so $a-V \ge 1$ so $a-1-V \ge 0$. So $a-a-b+1 \in S$.

Suppose $n \in S$ and $n > ab -b -a$.

Then $n = ax + by; x,y \ge 0$ wolog $0 \le x < b$. (Else substitute with $x' = x + kb$ so that $0 \le x < b$). $ax + by > a(b-1) + b(-1)$ but $x \le b-1$ so $y > -1$ i.e $y \ge 0$.

$n+1 = ax +by + aW -bV=a(x+W) + b(y-V) = a(x+W -b) + (y-V +a)$. $x+(W-b) > x \ge 0$. $y-V + a = y +(a-V) > y \ge 0$. So $n_1 \in S$.

So Claim 2 follows by induction.

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Example $a = 5$ and $b=7$. All integers $n > 35- 5 - y = 23$ are in $S$.

$3*5 - 2*7 = 1$ so $W =5; V = 2$ $24 ==a(W -1) + b(a-1-V)=5*2 + 7*2$. $25 = 5(2+3) + 7(2-2) = 5*5; 26 = 5(5+3) + 7(0-2)= 5(5+3 -7) + 7(0-2+5) = 5+3*7; $ etc.

So $S = \{0,5,10,15,20,7,12,17,22,14,19,21, n\ge 24\}$.