Common Divisor Divides Integer Combination/Corollary/Converse does not Hold

Theorem

Let $a, b, c \in \Z$ be integers.

Let:

$c \divides \paren {a + b}$

Then it is not necessarily the case that:

$c \divides a \land c \divides b$

Proof

Proof by Counterexample:

Let $a = 2, b = 4, c = 3$.

Then we have:

$3 \divides \paren {2 + 4}$

but:

$3 \nmid 2$

and:

$3 \nmid 4$

$\blacksquare$

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $3$