Intuition - Linear Congruence Theorem

Question

Let a and b be integers (not both 0) with greatest common divisor d.
Then an integer $c = ax + by$ for some $x, y \in Z$ $\iff d|c$.
In particular, d is the least positive integer of the form ax +by.

Is there intuition? Or illustration? I keep forgetting which variable is supposed to go. I'm not querying proofs.

Withal, if I write $ax + by = c$ as $ax \equiv c \; (mod \, b)$, then an error is even more likely! I can't remember if it's $c|d$, $b|c$, $c|a$, or some other wrong combination...

Origin - Elementary Number Theory, Jones, p10, Theorem 1.8

Answer 1

It's definitely something you can develop intuition for. Since $a$ and $b$ have common divison $d$, we have

$$c = ax+by = dzx+dwy = d(zx+dw)$$

for some $z,w$. Hence $d | c$. Think about it like simplifying fractions you learned in middle school or earlier. If everything in the numerator is a multiple of $d$, and the denominator is $d$, then the result is an integer.

Answer 2

Given $d=\gcd(a,b)$ Thus $a=d.k_1$ and $b=d.k_2$

Necessary part

Suppose $c=ax+by$ where $x,y\in Z$. Then $c=(dk_1)x+(dk_2)y$ i.e. $c=d(xk_1+yk_2)$

Thus $d|c$

Sufficient part

Suppose $d|c$. Let $c=d.k$

Since $gcd$ can be expressed as linear combination of the two numbers, there exist $x,y\in Z$ such that $d=ax+by$. Multiply with $k$ to get $c=dk=k(ax+by)=a(kx)+b(ky)=ax_0+by_0$

Answer 3

You have two integers $a$ and $b$ (not both $0$) with the greatest common divisor $d$. It means that $d|a$ and $d|b$. As we know, if $d|a$ and $d|b$ then $d|ax+by$ for all $x,y\in\mathbb{Z}$. Furthermore, one wonders what the set $$S=\{ax+by:x,y\in\mathbb{Z}\}.$$ Of course, $d$ divides each element in $S$ since is one of the common factor(s) between $a$ and $b$ which is equivalent to the necessary part of your statement.

The sufficient part of your statement is motivated by Bezout's_identity.