Proof

By hypothesis, $a > 0$ and $b > 0$ are co-prime positive integers.

Assume, $x$ runs through all values of a complete residue system modulo $a$ and $y$ runs through all values of a complete residue system modulo $b$.
By assumption, there are $ab$ numbers of the form $ax+by.$
If for two different pairs $(x_1,y_1),$ $(x_2,y_y)$ we have $(ax_1+by_1)(ab)\equiv (ax_2+by_2)(ab),$ then by congruence modulo a divisor we have $$\begin{array}{rcl} (ax_1+by_1)(b)&\equiv&(ax_2+by_2)(b)\\ (ax_1)(b)&\equiv&(ax_2)(b).\\ \end{array}$$
By cancellation of congruences with general factor, it follows $x_1(b)\equiv x_2(b).$
With a symmetrical argument, it follows $y_1(a) \equiv y_2(b).$
This means that among the $ab$ different numbers of the form $ax+by,$ all are incongruent.
In otherwords, $ab$ runs through all values of the complete residue system modulo $ab.$

Now, assume, $x$ runs through all values of a reduced residue system modulo $a$ and $y$ runs through all values of a complete residue system modulo $b$.
If $x$ and $b$ are not co-prime, then the greatest common divisor $\gcd(x,b) > 1.$
In this case, by divisibility laws $\gcd(x,b)\mid (ax+by)$ and $\gcd(x,b)\mid ab,$ therefore $\gcd(x,b)\mid \gcd(ax+by,ab).$
Altogether, from $1 < \gcd(x,b)$ it follows $\gcd(ax+by,ab) > 1.$
With a a symmetrical argument, it follows from $1 < \gcd(y,a)$ that $\gcd(ax+by,ab) > 1.$
With 1), it remains to be shown that if $1 = \gcd(x,b)$ and $1 = \gcd(y,a),$ then $\gcd(ax+by,ab) = 1.$
- Assume, a prime number is a divisor $p\mid \gcd(ax+by,ab).$
- Then $p$ would also divide $ab$ and without loss of generality $p\mid a.$
- Moreover, $p$ would also divide $ax+by,$ and therefore $p\mid by.$
- Finally, because $\gcd(a,b)$ it would follow $p\mid y.$
- But this is a contradiction to $\gcd(y,a)=1.$
Therefore $\gcd(ax+by,ab) = 1.$
∎

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