Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many find the use of imperative ("Find", "Show") to be rude when asking for help; please consider rewriting your post. – Arturo Magidin Apr 2 '12 at 19:08

4 Answers 4

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

share|cite|improve this answer

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).

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.