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Let $a$, $b$, $c$ be integers. Prove that if $a|b$ and $a|(b+c)$ then $a|c$.

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Hint: Try writing $c=(b+c)-b$

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Hint $\: $ If $\rm\:b \in a\:\mathbb Z\:$ then $\rm\:b+c\in a\:\mathbb Z \iff c\in a\:\mathbb Z\:$ by $\rm\:a\:\mathbb Z\:$ is closed under addition, subtraction.

Thus it is simply a divisibility translation of the fact that the set $\rm\:a\:\mathbb Z\:$ of multiples of $\rm\:a\:$ is closed under addition and subtraction, i.e. these multiples form a subgroup of $\mathbb Z$. The same holds true for the set $\rm M$ of common multiples of any finite subset $\rm\:S\subset \mathbb Z$. The $\mathbb Z$-linear structure of $\rm M,$ along with the Division Algorithm, implies the least positive element $\rm\:m\in M\:$ divides every element of $\rm M,$ so it is the least common multiple of $\rm S,\:$ i.e. $\rm\:S\ |\ n$ $\iff$ $\rm lcm\: S\ |\ n$ $\iff$ $\rm m\ |\ n,\:$ where $\rm\:S\ |\ n\:$ means $\rm\:s\ |\ n\:$ for all $\rm\:s\in S.\:$ From this follows immediately the existence of greatest common divisors (gcd), the prime divisor property, and the uniqueness of prime factorizations (fundamental theorem of arithmetic).

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HINT: Try $$b=aK_1 \tag{1}$$ $$(b+c) = aK_2 \tag{2}$$ (Note that $a \mid b \iff (1)$ and $a \mid (b+c) \iff (2)$)

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$ a|b & a|b+c b=a.k & b+c=a.h, where h,k are integers $ $\implies b+c-b=a(h-k), where \ h-k \ is \ also \ integer$ $\implies c=a.l ,where \ l=h-k$
$\implies a|c$

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