Divisor of Sum of Coprime Integers

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Theorem

Let $a, b, c \in \Z_{>0}$ such that:

$a \perp b$ and $c \backslash \left({a + b}\right)$.

where:

$a \perp b$ denotes $a$ and $b$ are coprime

$c \backslash \left({a + b}\right)$ denotes that $c$ is a divisor of $a + b$.

Then $a \perp c$ and $b \perp c$.

That is, a divisor of the sum of two coprime integers is coprime to both.

Let $d \in \Z_{>0}: d \backslash c \land d \backslash a$.

Then:

$\displaystyle $	$\displaystyle d$	$\backslash$	$\displaystyle \left({a + b}\right)$	$\displaystyle $	as $c \backslash \left({a + b}\right)$
$\displaystyle \implies$	$\displaystyle d$	$\backslash$	$\displaystyle \left({a + b - a}\right)$	$\displaystyle $
$\displaystyle \implies$	$\displaystyle d$	$\backslash$	$\displaystyle b$	$\displaystyle $
$\displaystyle \implies$	$\displaystyle d$	$=$	$\displaystyle 1$	$\displaystyle $	as $d \backslash a$ and $d \backslash b$ which are coprime

A similar argument shows that if $d \backslash c \land d \backslash b$ then $d \backslash a$.

It follows that:

$\gcd \left\{{a, c}\right\} = \gcd \left\{{b, c}\right\} = 1$

Hence the result.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle d\)	\(\backslash\)	\(\displaystyle \left({a + b}\right)\)	\(\displaystyle \)	as $c \backslash \left({a + b}\right)$
\(\displaystyle \implies\)	\(\displaystyle d\)	\(\backslash\)	\(\displaystyle \left({a + b - a}\right)\)	\(\displaystyle \)
\(\displaystyle \implies\)	\(\displaystyle d\)	\(\backslash\)	\(\displaystyle b\)	\(\displaystyle \)
\(\displaystyle \implies\)	\(\displaystyle d\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	as $d \backslash a$ and $d \backslash b$ which are coprime